Differentiation Under the Sign of Integration: Methods to Solve Integrals

Differentiation under the Sign of Integration: Integration is the branch of calculus that deals with finding the sum of small changes in a variable with respect to another variable. If we closely look at this definition, we can understand that integration is the reverse of derivative. While differentiation or taking derivative breaks down an area under a curve into small parts, integration sums together all those parts. Hence, the integral or integration of a function is also called antiderivative.

Differentiation or derivative is a branch of calculus that deals with finding the rate of change of a variable with respect to another variable. The slope of a curve has a similar definition. So, if we represent a function as a curve, then its slope is the derivative at that point. Derivative or differentiation of a variable \(y\) with respect to \(x\) is written as \(\frac{{dy}}{{dx}}\).

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Methods to Solve Integrals

Some integration problems involve functions that do not have antiderivatives. There are a few ways to solve such problems, which include,

1. Integration by Substitution

Sometimes, a well-chosen substitution will reduce a complex integration problem into a straightforward one.

Example:

Consider this problem, \(I = \int {\frac{{\log x}}{x}} dx\)

We do not have a formula that gives antiderivative of \(\log x\). Here we can substitute,

\(\log x = t\)

Differentiating both sides of this equation gives,

\(\frac{d}{{dx}}(\log x) = \frac{d}{{dx}}(t)\)

\( \Rightarrow \frac{1}{x} = \frac{{dt}}{{dx}}\)

\( \Rightarrow \frac{{dx}}{x} = dt\)

Making this substitution changes the integral to

\(\therefore I = \int {\frac{{dt}}{t}} \)

We know that the antiderivative of \(\frac{1}{t}\) is \(\log t\)

\( \Rightarrow I = \log t + c\)

\(\therefore I = \log (\log x) + c\)

2. Integration by Parts

Sometimes, the integrand is a product of two different functions from the types –

• Inverse (I)
• Logarithmic (L)
• Algebraic (A)
• Trigonometric(T)
• Exponential (E)

Of these functions, one is considered \(u\) and the other \(v\) based on the precedence order of ILATE.

The formula for integration by parts is,

\(\int u v\,dx = u\int v\,dx – \int {\left( {\int v\,dx} \right)} \left( {\frac{{du}}{{dx}}} \right)dx\)

Example:

Consider this problem, \(I = \int x \sin x\,dx\).

We know antiderivatives of \(x\) and also that of \(\sin x\), but we do not know the antiderivative of their product. Here we can use integration by parts. There are two functions – \(x\) which is algebraic and \(\sin x\) which is trigonometric. Hence, we assign \(x\) to be \(u\) and \(\sin x\) to be \(v\) according to the order ILATE.

\(I = \int x \sin x\,dx\)

\( \Rightarrow I = x\int {\sin }\,x\,dx – \int {\left( {\int {\sin }\, x\,dx} \right)} \left( {\frac{{d(x)}}{{dx}}} \right)dx\)

\( \Rightarrow I = x( – \cos x) – \int {( – \cos x)} (1)dx\)

\( \Rightarrow I = – x\cos x + \int {\cos x} \,dx\)

\(\therefore I = – x\cos x + \sin x + c\)

There are many more such methods that one can use according to the type of integrand.

3. Definite Integration Formulas

There are many easier ways to solve a definite integration. We can list down a couple of those,

\(\int_a^b f (x)dx = \int_a^b f (a + b – x)dx\)

\(\int_0^a f (x)dx = \int_0^a f (a – x)dx\)

When none of these methods works, one can try differentiation under the sign of integration.

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Differentiation Under the Sign of Integration

As the name suggests, this method involves differentiating under the integral sign. It is also called the Leibniz integral rule, named after Gottfried Leibniz. It is also called differentiation under the integral sign, DUIS in short.

This rule states that for an integral of the form, \(\int_{a(t)}^{b(t)} f (x,t)dx\), where \( – \infty < a(t),b(t) < \infty \), the derivative is given by,

\(\frac{d}{{dt}}\left( {\int_{a(t)}^{b(t)} f (x,t)dx} \right) = f(t,b(t))\frac{d}{{dt}}b(t) – f(t,a(t))\frac{d}{{dt}}a(t) + \int_{a(t)}^{b(t)} {\frac{\partial }{{\partial t}}} f(x,t)dx\)

The above equation shows the most general form of the Leibniz integral rule.

If \(a(t)\) and \(b(t)\) are constants, the special case of this rule is given by,

\(\frac{d}{{dt}}\left( {\int_a^b f (x,t)dx} \right) = \int_a^b {\frac{\partial }{{\partial t}}} f(x,t)dx\)

We will focus on the special case of the Leibniz integral rule.

We can think of this rule as going back a step behind to reach the goal. We take the derivative of the integral and find the answer to solve an integration problem.

Steps to Perform DUIS

Let \(I = \int_a^b f (x,t)dx\)

Here, the final answer \(I\) will be a function of the parameter \(t\).

Step 1: Differentiate both sides of the given equation with respect to \(t\) to get,

\(\frac{d}{{dt}}I = \frac{d}{{dt}}\int_a^b f (x,t)dx\)

\(\therefore \frac{{dI}}{{dt}} = \int_a^b {\frac{\partial }{{\partial t}}} f(x,t)dx\)

\(\therefore \frac{{dI}}{{dt}} = \int_a^b {{f^\prime }} (x,t)dx\)

Step 2: Integrate the RHS obtained in the above step. With the partial derivative taken in the first step, we can now do the integration by using the other methods we learned before. The answer obtained on RHS will be a function of \(t\) only.
\(\therefore \frac{{dI}}{{dt}} = g(t)\)

Step 3: Integrate both sides.

\(\frac{{dI}}{{dt}} = g(t)\)

\( \Rightarrow dI = g(t)dt\)

\( \Rightarrow \int d I = \int g (t)dt\)

\(\therefore I(t) = h(t) + c\)

Step 4: Substitute a value of \(t\), which will make the original integration \(0\) and thus, find the value of \(c\).

Solved Examples on Differentiation Under the Sign of Integration

Below are a few solved examples that can help in getting a better idea.

Q.1. Evaluate: \(\int_0^\infty {\frac{{{e^{ – x}}}}{x}} \left( {1 – {e^{ – \alpha x}}} \right)dx\)
Ans: In this problem, the independent variable is \(x\) and the parameter is \(\alpha \).
We start this problem by considering the given integration as \(I(\alpha )\)
\(\therefore I = I(\alpha ) = \int_0^\infty {\frac{{{e^{ – x}}}}{x}} \left( {1 – {e^{ – \alpha x}}} \right)dx\;\;\;\cdots (1)\)
Next, we apply DUIS to get,
\( \Rightarrow \frac{d}{{d\alpha }}I = \int_0^\infty {\frac{\partial }{{\partial \alpha }}} \frac{{{e^{ – x}}}}{x}\left( {1 – {e^{ – \alpha x}}} \right)dx\)
\( \Rightarrow \frac{{dl}}{{d\alpha }} = \int_0^\infty {\frac{{{e^{ – x}}}}{x}} \frac{\partial }{{\partial \alpha }}\left( {1 – {e^{ – \alpha x}}} \right)dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \int_0^\infty {\frac{{{e^{ – x}}}}{x}} \left( {\frac{\partial }{{\partial \alpha }}1 – \frac{\partial }{{\partial \alpha }}{e^{ – \alpha x}}} \right)dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \int_0^\infty {\frac{{{e^{ – x}}}}{x}} \left( {0 – {e^{ – \alpha x}}\frac{\partial }{{\partial \alpha }}( – \alpha x)} \right)dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \int_0^\infty {\frac{{{e^{ – x}}}}{x}} \left( { – {e^{ – \alpha x}}( – x)} \right)dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \int_0^\infty {\frac{{{e^{ – x}}{e^{ – \alpha x}}}}{x}} ( – ( – x))dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \int_0^\infty {{e^{ – (\alpha + 1)x}}} dx\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \left[ {\frac{{{e^{ – (\alpha + 1)x}}}}{{(\alpha + 1)}}} \right]_0^\infty \)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = \frac{{{e^{ – (\alpha + 1)\infty }}}}{{(\alpha + 1)}} – \frac{{{e^{ – (\alpha + 1)0}}}}{{(\alpha + 1)}}\)
\( \Rightarrow \frac{{dl}}{{d\alpha }} = \frac{{{e^{ – \infty }}}}{{(\alpha + 1)}} – \frac{{{e^0}}}{{(\alpha + 1)}}\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = 0 – \frac{1}{{(\alpha + 1)}}\)
\( \Rightarrow \frac{{dI}}{{d\alpha }} = – \frac{1}{{(\alpha + 1)}}\)
\( \Rightarrow dI = – \frac{1}{{(\alpha + 1)}}d\alpha \)
Integrating both sides we get,
\( \Rightarrow \int d I = \int – \frac{1}{{(\alpha + 1)}}d\alpha \)
\(\therefore I = – \log (\alpha + 1) + c\,\,\,\,\,\,\,….(2)\)
To find \(c\), we need to inspect the original equation and see which value of \(\alpha \) makes it \(0\).
If we substitute \(\alpha = 0\) in equation \(1\), we will get the value of \(I\) as \(0\).
Substituting \(\alpha = 0\) in equation \(2\) we get,
\( \Rightarrow 0 = – \log (0 + 1) + c\)
\( \Rightarrow 0 = – \log 1 + c\)
\( \Rightarrow 0 = 0 + c\)
\(\therefore c = 0\)
\( \Rightarrow I = – \log (\alpha + 1) + 0\)
\(\therefore I = – \log (\alpha + 1)\).

Q.2. Prove that \(\int_0^1 {\frac{{{x^t} – 1}}{{\log x}}} dx = \log (1 + t)\)
Ans: In this problem, the independent variable is \(x\) and the parameter is \(t\).
We start this problem by considering the given integration as \(I(t)\)
\(\therefore I = I(t) = \int_0^1 {\frac{{{x^t} – 1}}{{\log x}}} dx\,\,\,\,\,….(1)\)
Next, we apply DUIS to get,
\( \Rightarrow \frac{d}{{dt}}I = \int_0^1 {\frac{\partial }{{\partial t}}} \left( {\frac{{{x^t} – 1}}{{\log x}}} \right)dx\)
\( \Rightarrow \frac{{dI}}{{dt}} = \int_0^1 {\frac{1}{{\log x}}} \frac{\partial }{{\partial t}}\left( {{x^t} – 1} \right)dx\)
\( \Rightarrow \frac{{dI}}{{dt}} = \int_0^1 {\frac{1}{{\log x}}} \left( {\frac{\partial }{{\partial t}}{x^t} – \frac{\partial }{{\partial t}}1} \right)dx\)
\( \Rightarrow \frac{{dI}}{{dt}} = \int_0^1 {\frac{1}{{\log x}}} \left( {{x^t}\log x – 0} \right)dx\)
\( \Rightarrow \frac{{dI}}{{dt}} = \int_0^1 {\frac{1}{{\log x}}} {x^t}\log x\,dx\)
\(\therefore \frac{{dI}}{{dt}} = \int_0^1 {{x^t}} dx\)
Next, we can integrate the RHS to get,
\( \Rightarrow \frac{{dI}}{{dt}} = \left[ {\frac{{{x^t}}}{{t + 1}}} \right]_0^1\)
\( \Rightarrow \frac{{dI}}{{dt}} = \frac{{{1^t}}}{{t + 1}} – \frac{{{0^t}}}{{t + 1}}\)
\(\therefore \frac{{dI}}{{dt}} = \frac{1}{{t + 1}}\)
Next, we take the \(dt\) term on the RHS and integrate both sides.
\( \Rightarrow \int d I = \int {\frac{1}{{t + 1}}} dt\)
\(\therefore I = \log (t + 1) + c\,\,\,\,\,\,\,…..(2)\)
To find \(c\), we need to inspect the original equation and see which value of \(t\) makes it \(0\).
If we substitute \(t=0\) in equation \(1\), we will get the value of \(I\) as \(0\).
Substituting \(t=0\) in equation \(2\) we get,
\( \Rightarrow 0 = \log (0 + 1) + c\)
\( \Rightarrow 0 = \log 1 + c\)
\( \Rightarrow 0 = 0 + c\)
\(\therefore c = 0\)
\( \Rightarrow I = \log (t + 1) + 0\)
\(\therefore I = \log (t + 1)\).

Q.3. Evaluate \(\mathop \smallint \nolimits_0^\infty \frac{{{e^{ – ax}} – {e^{ – bx}}}}{x}dx\)
Ans:
In this problem, the independent variable is \(x\) and the parameters are \(a\) and \(b\).
We start this problem by considering the given integration as \(I(a)\). In this particular problem, we can also consider the integration as \(I(b)\). It would make no difference.
\(\therefore \,I = I\left( a \right) = \mathop \smallint \nolimits_0^\infty \frac{{{e^{ – ax}} – {e^{ – bx}}}}{x}dx\;\;\;\cdots (1)\)
Next, we apply DUIS to get,
\(\Rightarrow \frac{d}{{da}}I = \mathop \smallint \nolimits_0^\infty \frac{\partial }{{\partial a}}\left( {\frac{{{e^{ – ax}} – {e^{ – bx}}}}{x}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\frac{\partial }{{\partial a}}\left( {{e^{ – ax}} – {e^{ – bx}}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {\frac{\partial }{{\partial a}}{e^{ – ax}} – \frac{\partial }{{\partial a}}{e^{ – bx}}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {{e^{ – ax}}\frac{\partial }{{\partial a}}\left( { – ax} \right) – 0} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {{e^{ – ax}}\left( { – x} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = – \mathop \smallint \nolimits_0^\infty {e^{ – ax}}dx\)
\(\Rightarrow \frac{{dI}}{{da}} = – \left[ {\frac{{{e^{ – ax}}}}{{ – a}}} \right]_0^\infty \)
\(\Rightarrow \frac{{dI}}{{da}} = – \left( {\frac{{{e^{ – a\infty }}}}{{ – a}} – \frac{{{e^{ – a0}}}}{{ – a}}} \right)\)
\(\Rightarrow \frac{{dI}}{{da}} = – \left( {0 – \frac{1}{{ – a}}} \right)\)
\(\Rightarrow \frac{{dI}}{{da}} = – \frac{1}{a}\)
\(\therefore \,dI = – \frac{1}{a}da\)
Integrating both sides,
\(\smallint dI = – \smallint \frac{1}{a}da\)
\(\therefore \,I = – \log a + c\;\;\;\cdots (2)\)
To find \(c\), we need to inspect the original equation and see which value of \(a\) makes it \(0\).
If we substitute \(a = b\) in equation \(2\), we will get the value of \(I\) as \(0\).
Substituting \(a = b\) in equation \(2\) we get,
\(0 = – \log b + c\)
\(\therefore \,c = \log b\)
\(\Rightarrow I = – \log a + \log b\)
\(\Rightarrow I = \log b – \log a\)
\(\therefore \,I = \log \frac{b}{a}\).

Q.4. Evaluate \(\mathop \smallint \nolimits_0^\infty \frac{{{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) – {{\tan }^{ – 1}}\left( {\frac{x}{b}} \right)}}{x}dx\)
Ans:
In this problem, the independent variable is \(x\) and the parameters are \(a\) and \(b\).
We start this problem by considering the given integration as \(I(a)\). In this particular problem, we can also consider the integration as \(I(b)\). It would make no difference.
\(\therefore \,I = I\left( a \right) = \mathop \smallint \nolimits_0^\infty \frac{{{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) – {{\tan }^{ – 1}}\left( {\frac{x}{b}} \right)}}{x}dx\;\;\;\cdots (1)\)
Next, we apply DUIS to get,
\(\frac{d}{{da}}I = \mathop \smallint \nolimits_0^\infty \frac{\partial }{{\partial a}}\left( {\frac{{{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) – {{\tan }^{ – 1}}\left( {\frac{x}{b}} \right)}}{x}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\frac{\partial }{{\partial a}}\left( {{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) – {{\tan }^{ – 1}}\left( {\frac{x}{b}} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {\frac{\partial }{{\partial a}}{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) – \frac{\partial }{{\partial a}}{{\tan }^{ – 1}}\left( {\frac{x}{b}} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {\frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}}\frac{\partial }{{\partial a}}\left( {\frac{x}{a}} \right) – 0} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {\frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}}x\frac{\partial }{{\partial a}}\left( {\frac{1}{a}} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{x}\left( {\frac{x}{{\frac{{{a^2} + {x^2}}}{{{a^2}}}}}\left( { – \frac{1}{{{a^2}}}} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = – \mathop \smallint \nolimits_0^\infty \frac{1}{{{a^2} + {x^2}}}dx\)
\(\Rightarrow \frac{{dI}}{{da}} = – \left[ {\frac{1}{a}{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right)} \right]_0^\infty \)
\(\Rightarrow \frac{{dI}}{{da}} = – \frac{1}{a}\left[ {{{\tan }^{ – 1}}\left( {\frac{\infty }{a}} \right) – {{\tan }^{ – 1}}\left( {\frac{0}{a}} \right)} \right]\)
\(\Rightarrow \frac{{dI}}{{da}} = – \frac{1}{a}\left[ {{{\tan }^{ – 1}}\left( \infty \right) – {{\tan }^{ – 1}}\left( 0 \right)} \right]\)
\(\Rightarrow \frac{{dI}}{{da}} = – \frac{1}{a}\left[ {\frac{\pi }{2} – 0} \right]\)
\(\Rightarrow \frac{{dI}}{{da}} = – \frac{\pi }{{2a}}\)
\(\therefore \,dI = – \frac{\pi }{{2a}}da\)
Integrating both sides,
\(\smallint dI = \smallint – \frac{\pi }{{2a}}da\)
\(\Rightarrow I = – \frac{\pi }{2}\smallint \frac{1}{a}da\)
\(\therefore \,I = – \frac{\pi }{2}(\log a) + c\;\;\;\cdots (2)\)
To find \(c\), we need to inspect the original equation and see which value of \(a\) makes it \(0\).
If we substitute \(a = b\) in equation \(1\), we will get the value of \(I\) as \(0\).
Substituting \(a = b\) in equation \(2\) we get,
\(0 = – \frac{\pi }{2}(\log b) + c\)
\(\therefore \,c = \frac{\pi }{2}(\log b)\)
\(\Rightarrow I = – \frac{\pi }{2}(\log a) + \frac{\pi }{2}(\log b)\)
\(\Rightarrow I = \frac{\pi }{2}\left[ { – \log a + \log b} \right]\)
\(\Rightarrow I = \frac{\pi }{2}\left[ {\log b – \log a} \right]\)
\(\therefore \,I = \frac{\pi }{2}\log \frac{b}{a}\).

Q.5. Evaluate \(\mathop \smallint \nolimits_0^\infty \frac{{{e^x} – {e^{ – ax}}}}{{x\,\sec \,x}}dx\)
Ans:
In this problem, the independent variable is \(x\) and the parameter is \(a\).
We start this problem by considering the given integration as \(I(a)\)
\(\therefore \,I = I\left( a \right) = \mathop \smallint \nolimits_0^\infty \frac{{{e^x} – {e^{ – ax}}}}{{x\,\sec \,x}}dx\;\;\;\cdots (1)\)
Next, we apply DUIS to get,
\(\frac{d}{{da}}I = \mathop \smallint \nolimits_0^\infty \frac{\partial }{{\partial a}}\left( {\frac{{{e^x} – {e^{ – ax}}}}{{x\sec x}}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{{x\sec x}}\frac{\partial }{{\partial a}}\left( {{e^x} – {e^{ – ax}}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{{x\sec x}}\left( {\frac{\partial }{{\partial a}}{e^x} – \frac{\partial }{{\partial a}}{e^{ – ax}}} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{{x\sec x}}\left( {0 – {e^{ – ax}}\frac{\partial }{{\partial a}}\left( { – ax} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{1}{{x\sec x}}\left( {0 – {e^{ – ax}}\left( { – x} \right)} \right)dx\)
\(\Rightarrow \frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty \frac{{{e^{ – ax}}}}{{\sec x}}dx\)
\(\therefore \,\frac{{dI}}{{da}} = \mathop \smallint \nolimits_0^\infty {e^{ – ax}}\cos \,x\,dx\;\;\;\cdots (2)\)
We can solve the RHS by using integration by parts. By ILATE order, we will consider \(\cos x\) as \(u\) and \(e^{-ax}\) as \(v\).
Let’s solve the RHS separately and call it \(K\).
\(K = \mathop \smallint \nolimits_0^\infty {e^{ – ax}}\cos \,x\,dx\)
\(\Rightarrow K = \cos \,x\smallint {e^{ – ax}}dx – \smallint \left( {\smallint {e^{ – ax}}dx} \right)\left( {\frac{d}{{dx}}\cos \,x} \right)dx\)
\(\Rightarrow K = \cos \,x\frac{{{e^{ – ax}}}}{{ – a}} – \smallint \frac{{{e^{ – ax}}}}{{ – a}}\left( { – \sin \,x} \right)dx\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\smallint {e^{ – ax}}\sin \,x\,dx\)
We need to apply integration by parts one more time.
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\left[ {\sin \,x\smallint {e^{ – ax}}dx – \smallint \left( {\smallint {e^{ – ax}}dx} \right)\left( {\frac{d}{{dx}}\sin \,x} \right)dx} \right]\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\left[ {\frac{{{e^{ – ax}}\sin \,x}}{{ – a}} – \smallint \frac{{{e^{ – ax}}}}{{ – a}}\cos \,x\,dx} \right]\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\left[ {\frac{{{e^{ – ax}}\sin \,x}}{{ – a}} + \frac{1}{a}\smallint {e^{ – ax}}\cos \,x\,dx} \right]\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\left[ {\frac{{{e^{ – ax}}\sin \,x}}{{ – a}} + \frac{1}{a}K} \right]\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} – \frac{1}{a}\left[ {\frac{{{e^{ – ax}}\sin \,x}}{{ – a}} + \frac{K}{a}} \right]\)
\(\Rightarrow K = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} + \frac{{{e^{ – ax}}\sin \,x}}{{{a^2}}} – \frac{K}{{{a^2}}}\)
\(\Rightarrow K + \frac{K}{{{a^2}}} = \frac{{{e^{ – ax}}\cos \,x}}{{ – a}} + \frac{{{e^{ – ax}}\sin \,x}}{{{a^2}}}\)
\(\Rightarrow K\left( {1 + \frac{1}{{{a^2}}}} \right) = \frac{{{e^{ – ax}}}}{a}\left( { – \cos \,x + \frac{{\sin \,x}}{a}} \right)\)
\(\Rightarrow K\left( {\frac{{{a^2} + 1}}{{{a^2}}}} \right) = \frac{{{e^{ – ax}}}}{a}\left( {\frac{{ – a\,\cos \,x + \sin \,x}}{a}} \right)\)
\(\Rightarrow K\left( {{a^2} + 1} \right) = {e^{ – ax}}\left( { – a\,\cos \,x + \sin \,x} \right)\)
\(\Rightarrow K = \left[ {\frac{{{e^{ – ax}}\left( { – a\,\cos \,x + \sin \,x} \right)}}{{\left( {{a^2} + 1} \right)}}} \right]_0^\infty \)
\(\Rightarrow K = \frac{1}{{\left( {{a^2} + 1} \right)}}\left( {{e^{ – a\infty }}\left( { – a\,\cos \,\infty + \sin \,\infty } \right) – {e^{ – a0}}\left( { – a\,\cos \,0 + \sin \,0} \right)} \right)\)
\(\Rightarrow K = \frac{1}{{\left( {{a^2} + 1} \right)}}\left( {0\left( { – a\,\cos \,\infty + \sin \,\infty } \right) – 1\left( { – a + 0} \right)} \right)\)
\(\Rightarrow K = \frac{1}{{\left( {{a^2} + 1} \right)}}\left( {0 + a} \right)\)
\(\therefore \,K = \frac{a}{{\left( {{a^2} + 1} \right)}}\)
\(\Rightarrow \frac{{dI}}{{da}} = \frac{a}{{\left( {{a^2} + 1} \right)}}\)
\(\therefore \,dI = \frac{a}{{\left( {{a^2} + 1} \right)}}da\)
Integrating both sides we get,
\(\smallint dI = \smallint \frac{a}{{\left( {{a^2} + 1} \right)}}da\)
\(\Rightarrow I = \frac{1}{2}\smallint \frac{{2a}}{{\left( {{a^2} + 1} \right)}}da\)
\(\therefore \,I = \frac{1}{2}\log \left( {{a^2} + 1} \right) + c\)
To find \(c\), we need to inspect the original equation and see which value of \(a\) makes it \(0\).
If we substitute \(a = – 1\) in equation \(1\), we will get the value of \(I\) as \(0\).
Substituting \(a = – 1\) in equation \(2\) we get,
\(0 = \frac{1}{2}\log \left( {{{\left( { – 1} \right)}^2} + 1} \right) + c\)
\(\Rightarrow 0 = \frac{1}{2}\log \left( {1 + 1} \right) + c\)
\(\Rightarrow 0 = \frac{1}{2}\log 2 + c\)
\(\therefore \,c = – \frac{1}{2}\log 2\)
\(\Rightarrow I = \frac{1}{2}\log \left( {{a^2} + 1} \right) – \frac{1}{2}\log 2\)
\(\Rightarrow I = \frac{1}{2}\left( {\log \left( {{a^2} + 1} \right) – \log 2} \right)\)
\(\therefore \,I = \frac{1}{2}\log \frac{{{a^2} + 1}}{2}\).

Summary

Differentiation or derivative and integration are two essential parts of calculus. A derivative is the rate of change of a dependent variable with a change in an independent variable. Integration involves adding small changes in a variable over an interval to find the total change in the variable.

In short, a derivative makes small parts out of a whole, and integration adds many small parts to create a whole. Thus, integration is the opposite of derivative, and hence, integration is also called antiderivative. There are two types of integrations – indefinite and definite. One of the most important methods to solve an integration is differentiation under the integral sign (DUIS).

Frequently Asked Questions (FAQs)

Students might be having many questions regarding Differentiation Under the Sign of Integration. Here are a few commonly asked questions and answers.

Q.1. How do you do the differentiation under the integral sign?
Ans: We do differentiation under the integral sign by taking partial derivatives.

Q.2. When can you move a derivative inside an integral?
Ans: Derivatives can be moved inside an integral when the limits of integration are independent of the parameter present in the integrand.

Q.3. How do you use the Leibniz rule?
Ans: Leibniz rule is used to find the integration of a function of a variable and a parameter by taking differentiation under the integral sign.

Q.4. What is the Newton Leibniz forumla?
Ans: The Leibniz formula is also known as the Newton Leibniz formula. This rule states that for an integral of the form,
\(\mathop \smallint \nolimits_{a\left( t \right)}^{b\left( t \right)} f\left( {x,t} \right)dx\), where \( – \infty < a\left( t \right),b\left( t \right) < \infty \) the derivative of this integral is given by the formula,
\(\frac{d}{{dt}}\left( {\mathop \smallint \nolimits_{a\left( t \right)}^{b\left( t \right)} f\left( {x,t} \right)dx} \right) = f\left( {t,\;b\left( t \right)} \right)\frac{d}{{dt}}b\left( t \right) – f\left( {t,\;a\left( t \right)} \right)\frac{d}{{dt}}a\left( t \right) + \mathop \smallint \nolimits_{a\left( t \right)}^{b\left( t \right)} \frac{\partial }{{\partial t}}f\left( {x,t} \right)dx\)
The above equation shows the most general form of the Leibniz integral rule.
If \(a(t)\) and \(b(t)\) are constants, then the special case of this rule is given by,
\(\frac{d}{{dt}}\left( {\mathop \smallint \nolimits_a^b f\left( {x,t} \right)dx} \right) = \mathop \smallint \nolimits_a^b \frac{\partial }{{\partial t}}f\left( {x,t} \right)dx\).

Q.5. What is the integration formula?
Ans: Integration formulas are a set of formulae for certain standard functions which give their antiderivatives. Few of them are listed below.
\(\smallint \cos \,x\,dx = \sin \,x + c\)
\(\smallint \sin \,x\,dx = – \cos \,x + c\)
\(\smallint \frac{1}{x}dx = \log \,x + c\)
\(\smallint {e^x}dx = {e^x} + c\).

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