Integral of Trigonometric Functions: If we know an object’s instantaneous velocity at a given time, a logical issue arises: can we calculate the object’s location at any given time? There are various practical & theoretical instances or scenarios involving the integration process. 

The expansion of integral calculus results from attempting to solve the problem of finding a function whenever its derivative is provided. It also results from the problem of finding the area enclosed by the graph of a function under specific conditions. These two problems result in two types of integrals, indefinite and definite integrals, making up the Integral Calculus. In this article, we will learn some standard formulae of trigonometric functions, and discuss the techniques to evaluate various forms of integrals involving trigonometric functions.

Primitive or Anti-derivative

A function \(\phi \left( x \right)\) is called a primitive (or an anti-derivative or an integral) of a function \(f\left( x \right)\) if \({\phi ^\prime }\left( x \right) = f\left( x \right).\)
For example, \(\frac{{{x^4}}}{4}\) is a primitive of \({x^3},\) because \(\frac{d}{{dx}}\left( {\frac{{{x^4}}}{4}} \right) = {x^3}.\)

Indefinite Integral

Consider the function \(f\left( x \right).\) The indefinite integral of \(f\left( x \right)\) is thus represented as \(\int f \left( x \right)dx,\) which is the family of all its primitives (or anti-derivatives).
The symbol \(\int f \left( x \right)dx\) is read as the indefinite integral of \(f\left( x \right)\) with respect to \(x.\)
Thus, \(\frac{d}{{dx}}\left( {\phi \left( x \right) + C} \right) = f(x) \Leftrightarrow \int f \left( x \right)dx = \phi \left( x \right) + C\) where \(C\) is an arbitrary constant known as the constant of integration and \(\phi \left( x \right)\) is primitive of \(f\left( x \right).\)
Here, the integral sign \(\int {,f} \left( x \right)\) is the integrand, the integration variable is \(x\) and the differential of \(x\) is \(dx.\)

Learn Everything About Trigonomtery Here

Fundamental Integration Formulas of Trigonometric Functions

Since \(\frac{d}{{dx}}\left[ {\phi \left( x \right)} \right] = f\left( x \right) \Leftrightarrow \int f \left( x \right)dx = \phi \left( x \right) + C.\)
Based upon this and various standard differentiation formulae, we obtain the following integration formulae of trigonometric functions:
1. \(\int {\sin } \,x\,dx = \, – \,\cos x + C\)
2. \(\int {\cos } \,x\,dx = \sin x + C\)
3. \(\int {{{\sec }^2}} x\,dx = \tan x + C\)
4. \(\int {{{\operatorname{cosec} }^2}} x\,dx = – \cot x + C\)
5. \(\int {\sec } \,x\tan x\,dx = \sec x + C\)
6. \(\int {\operatorname{cosec} } \,x\cot x\,dx = – \operatorname{cosec} x + C\)
7. \(\int {\cot } \,x\,dx = \log \left| {\sin x} \right| + C\)
8. \(\int {\sec } \,x\,dx = \log \left| {\sec x + \tan x} \right| + C\)
9. \(\int {\operatorname{cosec} } \,x\,dx = \log \left| {\operatorname{coses} x – \cot x} \right| + C\)

Integral of the Form: \(\int f \left( {ax + b} \right)dx\)

If \(\int f \left( x \right)dx = \phi \left( x \right),\) then \(\int f \left( {ax + b} \right)dx = \frac{1}{a}\phi \left( {ax + b} \right)\)
We can prove the above result as follows:
Step 1: Let \(I = \int f \left( {ax + b} \right)dx\)
Step 2: Substitute \(ax + b = t \Rightarrow dx = \frac{1}{a}dt.\)
Step 3: \(I = \int f \left( {ax + b} \right)dx = \frac{1}{a}\int f \left( t \right)dt\)
Step 4: \(I = \frac{1}{a}\int f \left( t \right)dt = \frac{1}{a}\phi \left( t \right)\) [Since \(\int f \left( x \right)dx = \phi \left( x \right)\)].
Step 5: Substitute back the value of \(t,\) So, we have
\(I = \int f \left( {ax + b} \right)dx = \frac{1}{a}\phi \left( {ax + b} \right)\)
Based on the above result, we have some standard formulae of trigonometric functions
1. \(\int {\sin } \left( {ax + b} \right)dx = – \frac{1}{a}\,\cos \left( {ax + b} \right) + C\)
2. \(\int {\cos } \left( {ax + b} \right)dx = \frac{1}{a}\,\sin \left( {ax + b} \right) + C\)
3. \(\int {{{\sec }^2}} \left( {ax + b} \right)dx = \frac{1}{a}\,\tan \left( {ax + b} \right) + C\)
4. \(\int {{{\operatorname{cosec} }^2}} \left( {ax + b} \right)dt = – \frac{1}{a}\,\cot \left( {ax + b} \right) + C\)
5. \(\int {\sec } \left( {ax + b} \right)\tan \left( {ax + b} \right)dx = \frac{1}{a}\,\sec \left( {ax + b} \right) + C\)
6. \(\int {\operatorname{cosec} } \left( {ax + b} \right)\cot \left( {ax + b} \right)dx = – \frac{1}{a}\,\operatorname{cosec} \left( {ax + b} \right) + C\)
7. \(\int {\tan } \left( {ax + b} \right)dx = – \frac{1}{a}\,\log \left| {\cos \left( {ax + b} \right)} \right| + C\)
8. \(\int {\cot } \left( {ax + b} \right)dx = \frac{1}{a}\,\log \left| {\sin \left( {ax + b} \right)} \right| + C\)
9. \(\int {\sec } \left( {ax + b} \right)dx = \frac{1}{a}\,\log \left| {\sec \left( {ax + b} \right) + \tan \left( {ax + b} \right)} \right| + C\)
10. \(\int {\operatorname{cosec} } \left( {ax + b} \right)dx = \frac{1}{a}\,\log \left| {\operatorname{cosec} \left( {ax + b} \right) – \cot \left( {ax + b} \right)} \right| + C\)

Evaluation of Integrals of the Form: \(\int {{{\sin }^m}}\,xdx,\int {{{\cos }^m}}\,xdx,\) where \(m \leqslant 4,\,m \in N\)

To evaluate integrals of the form \(\int {{{\sin }^m}}\,xdx,\int {{{\cos }^m}}\,xdx,\) where \(m \leqslant 4\), we express \({\sin ^m}x\) and \({\cos ^m}x\) in terms of sines and cosines of multiples of \(x\) by using the following trigonometrical identities:
1. \({\sin ^2}x = \frac{{1 – \cos 2x}}{2}\)
2. \({\cos ^2}x = \frac{{1 + \cos 2x}}{2}\)
3. \(\sin 3x = 3\sin x – 4\,{\sin ^3}\,x\)
4. \(\cos 3x = 4\,{\cos ^3}\,x – 3\cos x\)

Evaluation of Integrals of the Form: \(\int {\sin mx\cos nxdx} ,\) \(\int {\sin mx\sin nxdx} ,\)
\(\int {\cos mx\cos nxdx} \)

To evaluate these integrals, we use the following trigonometrical identities to express the products into sums.
1. \(2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A – B} \right)\)
2. \(2\cos A\sin B = \sin \left( {A + B} \right) – \sin \left( {A – B} \right)\)
3. \(2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A – B} \right)\)
4. \(2\sin A\sin B = \cos \left( {A – B} \right) – \cos \left( {A + B} \right)\)

Integrals of the Form: \(\int {\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}}} dx = \log \left| {f\left( x \right)} \right| + C\)

Step 1: Let \(I = \int {\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}}} dx.\)
Step 2: Substitute \(f\left( x \right) = t \Rightarrow {f^\prime }\left( x \right)dx = dt\)
Step 3: \(I = \int {\frac{{{f^\prime }\left( x \right)}}{{f\left( x \right)}}} dx = \int {\frac{1}{t}} \,dt = \log \left| t \right| + C = \log \left| {f\left( x \right)} \right| + C.\)

Note: If the numerator in the integrand is the exact differential of the denominator, then its integral is the logarithm of the denominator.

Some Standard Results

1. \(\int {\tan } \,x\,dx = – \log \left| {\cos x} \right| + C\)
2. \(\int {\cot } \,x\,dx = \log \left| {\sin x} \right| + C\)
3. \(\int {\sec } \,x\,dx = \log \left| {\sec x + \tan x} \right| + {\text{C}}.\)
4. \(\int {\operatorname{cosec} } \,x\,dx = \log \left| {\operatorname{cosec} x – \cot x} \right| + C\)

Integrals of the Form: \(\int {{{\tan }^m}\,x\,{\text{se}}{{\text{c}}^{2n}}\,xdx} ,\) \(\int {{{\cot }^m}\,x\,{\text{cose}}{{\text{c}}^{2n}}\,xdx} ;\)
\(m,\,n \in N\)

In order to evaluate this type of integral. We may follow the following algorithm.
Step 1: Write the given integral as \(I = \int {{{\tan }^m}}\,x{\left( {{{\sec }^2}\,x} \right)^{\left( {n – 1} \right)}}{\sec ^2}\,x\,dx\)
Step 2: Put \(\tan x = t \Rightarrow {\sec ^2}\,x\,dx = dt\) and write the integral as
\(I = \int {{{\tan }^m}}\,x{\left( {{{\sec }^2}\,x} \right)^{n – 1}}\,{\sec ^2}\,x\,dx\)
\( = \int {{{\tan }^m}}\,x{\left( {1 + {{\tan }^2}\,x} \right)^{n – 1}}\,{\sec ^2}\,x\,dx\)
\( = \int {{t^m}} {\left( {1 + {t^2}} \right)^{n – 1}}\,dt\)
Step 3: Expand \({\left( {1 + {t^2}} \right)^{n – 1}}\) by binomial theorem in step \(2\) and integrate.
Step 4: Replace \(t\) by \(\tan x\) in step \(3.\)

Integrals of the Form: \(\int {{{\tan }^n}}\,x\,dx,\int {{{\cot }^n}}\,x\,dx,\) where \(n \in N,\,n \leqslant 4\)

To evaluate the integral of this form, we use the following trigonometric identities:
1. \({\tan ^2}\,x = {\sec ^2}\,x – 1\)
2. \({\cot ^2}\,x = {\operatorname{cosec} ^2}\,x – 1\)

Integrals of the Form: \(\int {{{\sin }^m}}\,x\,{\cos ^m}\,x\,dx,\) where \(m,n \in N\)

Algorithm:
Step 1: Obtain the integral, say, \(\int {{{\sin }^m}}\,x\,{\cos ^n}\,x\,dx\)
Step 2: Check the exponents of \(\sin \,x\) and \(\cos \,x.\)
Step 3: Follow the following steps:
i. If the exponent of \(\sin \,x\) is an odd integer put \(\cos \,x = t.\)
ii. If the exponent of \(\cos \,x\) is an odd integer put \(\sin \,x = t.\)
iii. If the exponents of \(\sin \,x\) and \(\cos \,x\) both are odd positive integers put either \(\sin \,x = t\) or \(\cos \,x = t.\)
iv. If the exponents of \(\sin \,x\) and \(\cos \,x.\) both are even positive integers then express \({\sin ^m}\,x\,{\cos ^n}x\) in terms of sines and cosines of multiples of \(x\) by using trigonometric results.
Step 5: Evaluate the integral obtained in step \(3.\)

Evaluate Integrals of the form \(\int {{{\sin }^m}} x\,{\cos ^n}x\,dx\) where \(m,n \in Q\) and \(m + n\) is a negative integer:

Algorithm:
Step 1: Change the integrand in terms of \(\tan x\) and \({\sec ^2}\,x.\)
Step 2: Divide numerator and denominator by \({\cos ^k}x,\) where \(k = – \left( {m + n} \right)\)
Step 3: Substitute \(\tan x = t\) and use the standard formulae

Evaluation of Integrals by Using the Trigonometric Substitutions

Expression	Substitution
\({a^2} + {x^2}\)	\(x = a\tan \theta \) or \(a\cot \theta \)
\({a^2} – {x^2}\)	\(x = a\sin \theta \) or \(a\cos \theta \)
\({x^2} – {a^2}\)	\(x = a\sec \theta \) or \(a\operatorname{cosec} \theta \)
\(\sqrt {\frac{{a – x}}{{a + x}}} \) or, \(\sqrt {\frac{{a + x}}{{a – x}}} \)	\(x = a\cos 2\theta \)
\(\sqrt {\frac{{x – \alpha }}{{\beta – x}}} \) or \(\sqrt {\left( {x – \alpha } \right)\left( {x – \beta } \right)} \)	\(x = \alpha\,{\cos ^2}\theta + \beta\,{\sin ^2}\theta .\)

Integration Using Trigonometric Identities

While integrating a function with any form of trigonometric integrands, we employ trigonometric identities to simplify the functions.

Integrals of the Form, \(\int {\frac{1}{{a\sin x + b\cos x}}} dx,\int {\frac{1}{{a + b\sin x}}} dx\)
\(\int {\frac{1}{{a + b\cos x}}} dx,\int {\frac{1}{{a\sin x + b\cos x + c}}} dx\)

Step 1: Put \(\sin x = \frac{{2\tan\,\frac{x}{2}}}{{1 + {{\tan }^2}\,\frac{x}{2}}}\) and \(\cos x = \frac{{1 – {{\tan }^2}\,\frac{x}{2}}}{{1 + {{\tan }^2}\,\frac{x}{2}}}\) and simplify.
step 2: Replace \(1 + {\tan ^2}\,\frac{x}{2}\) in the numerator by \({\sec ^2}\,\frac{x}{2}.\)
Step 3: \(\tan\,\frac{x}{2} = t \Rightarrow \frac{1}{2}{\sec ^2}\,\frac{x}{2}dx = dt.\) This substitution reduces the integral in the form \(\int {\frac{1}{{a{t^2} + bt + c}}} dt.\)
Step 4: Evaluate integral obtained in step \(3\) by converting it into the standard form whose formula is known to us.

Integrals of the Form, \(\int {\frac{1}{{a\,{{\sin }^2}\,x + b\,{{\cos }^2}\,x}}} dx,\int {\frac{1}{{a + b\,{{\sin }^2}\,x}}} dx\)
\(\int {\frac{1}{{a + b\,{{\cos }^2}\,x}}} dx,\int {\frac{1}{{{{\left( {a\sin x + b\cos x} \right)}^2}}}} dx\)

To evaluate these types of integrals we use the following algorithm.
Algorithm:
Step 1: Divide numerator and denominator both by \({\cos ^2}x.\)
Step 2: Replace \({\sec ^2}x,\) by \(1 + {\tan ^2}x\)
Step 3: Put \(\tan x = t\) so that \({\sec ^2}\,xdx = dt.\) This substitution reduces the integral in the form, \(\int {\frac{1}{{a{t^2} + bt + c}}} dt\)
Step 4: Evaluate integral obtained in step \(3\) by converting it into the standard form whose formula is known to us.

Reduction Formulae

A reduction formula is frequently used to calculate integrals of higher powers during integration. By applying this formula, we can reduce the power of trigonometric functions until its integration can be evaluated.

1. \(\int {{{\sin }^n}}\,x\,dx = \frac{{ – \cos x \cdot {{\sin }^{n – 1}}\,x}}{n} + \frac{{n – 1}}{n}\int {{{\sin }^{n – 2}}}\,x\,dx\)
2. \(\int {{{\cos }^n}}\,x\,dx = \frac{{\sin x \cdot {{\cos }^{n – 1}}\,x}}{n} + \frac{{n – 1}}{n}\int {{{\cos }^{n – 2}}}\,x\,dx\)
3. \(\int {{{\tan }^n}}\,x\,dx = \frac{{{{\tan }^{n – 1}}\,x}}{{n – 1}} – \int {{{\tan }^{n – 2}}}\,x\,dx,n \geqslant 2\)
4. \(\int {{{\cot }^n}}\,x\,dx = – \frac{{{{\cot }^{n – 1}}\,x}}{{n – 1}} – {I_{n – 2}},n \geqslant 2\)
5. \(\int {{{\sec }^n}}\,x\,dx = \frac{{\tan x\,{{\sec }^{n – 2}}\,x}}{{n – 1}} + \frac{{n – 2}}{{n – 1}}{I_{n – 2}}\)

Solved Examples – Integral of Trigonometric Functions

Q.1. Evaluate : \(\int {\frac{1}{{{a^2}\,{{\sin }^2}\,x + {b^2}\,{{\cos }^2}\,x}}} dx\)
Ans: \(\int {\frac{1}{{{a^2}\,{{\sin }^2}\,x + {b^2}\,{{\cos }^2}\,x}}} dx\)
Dividing the numerator and denominator of the given integrand by \({\cos ^2}\,x,\) we get
\(I = \int {\frac{1}{{{a^2}\,{{\sin }^2}\,x + {b^2}\,{{\cos }^2}\,x}}} dx = \int {\frac{{{{\sec }^2}\,x}}{{{a^2}\,{{\tan }^2}\,x + {b^2}}}} dx\)
Putting \(\tan x = t\) and \({\sec ^2}x\,dx = dt,\) we get
\(I = \int {\frac{{dt}}{{{a^2}{t^2} + {b^2}}}} \)
\( = \frac{1}{{{a^2}}}\int {\frac{{dt}}{{{t^2} + {{\left( {\frac{b}{a}} \right)}^2}}}} \)
\( = \frac{1}{{{a^2}}} \times \frac{1}{{\frac{b}{a}}}\,{\tan ^{ – 1}}\left( {\frac{t}{{\frac{b}{a}}}} \right) + C\)
\( = \frac{1}{{ab}}\,{\tan ^{ – 1}}\left( {\frac{{at}}{b}} \right) + C\)
\( = \frac{1}{{ab}}\,{\tan ^{ – 1}}\left( {\frac{{a\tan x}}{b}} \right) + C\)
Hence, \(\int {\frac{1}{{{a^2}\,{{\sin }^2}\,x + {b^2}\,{{\cos }^2}\,x}}} dx = \frac{1}{{ab}}\,{\tan ^{ – 1}}\left( {\frac{{a\tan x}}{b}} \right) + C\)

Q.2. Evaluate: \(\int {{{\sin }^3}}\,x\,{\cos ^4}\,x\,dx\)
Ans: \(I = \int {{{\sin }^3}}\,x\,{\cos ^4}\,x\,dx\)
Here,power of \(\sin \,x\) is odd, so we substitute \(\cos \,x = t\)
\( \Rightarrow – \sin xdx = dt \Rightarrow dx = – \frac{{dt}}{{\sin x}}\)
\(\therefore I = \int {{{\sin }^3}} x{t^4}\left( { – \frac{{dt}}{{\sin x}}} \right)\)
\( = – \int {\left( {{{\sin }^2}x \times {t^4}} \right)} dt\)
\( = – \int {\left( {1 – {t^2}} \right)} {t^4}dt\)
\( = – \int {\left( {{t^4} – {t^6}} \right)} dt\)
\( \Rightarrow I = – \frac{{{t^5}}}{5} + \frac{{{t^7}}}{7} + C\)
Hence, \(\int {{{\sin }^3}}\,x\,{\cos ^4}\,x\,dx = – \frac{{{{\cos }^5}\,x}}{5} + \frac{{{{\cos }^7}\,x}}{7} + C\)

Q.3. Evaluate: \(\int {\frac{{{{\sin }^4}\,x}}{{{{\cos }^8}\,x}}} dx\)
Ans: \(I = \int {\frac{{{{\sin }^4}\,x}}{{{{\cos }^8}\,x}}} dx\)
\( = \int {\left( {\frac{{\frac{{{{\sin }^4}\,x}}{{{{\cos }^4}\,x}}}}{{\frac{{{{\cos }^8}\,x}}{{{{\cos }^4}\,x}}}}} \right)} dx\)
\( = \int {\left( {{{\tan }^4}\,x\,{{\sec }^4}\,x} \right)} dx\)
\( = \int {\left\{ {{{\tan }^4}\,x\left( {1 + {{\tan }^2}\,x} \right){{\sec }^2}\,x} \right\}} dx\)
Putting \(\tan \,x = t,\) and \({\sec ^2}\,x\,dx = dt\) we get
\(I = \int {{t^4}} \left( {1 + {t^2}} \right)dt\)
\( = \frac{{{t^5}}}{5} + \frac{{{t^7}}}{7} + C\)
Substituting for \(t,\) we get
\(\therefore I = \frac{{{{\tan }^5}\,x}}{5} + \frac{{{{\tan }^7}\,x}}{7} + C\)
Hence, \(\int {\frac{{{{\sin }^4}\,x}}{{{{\cos }^8}\,x}}} dx = \frac{{{{\tan }^5}\,x}}{5} + \frac{{{{\tan }^7}\,x}}{7} + C\)

Q.4. Evaluate: \(\int {\frac{{{{\sin }^8}\,x – {{\cos }^8}\,x}}{{1 – 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x}}} dx\)
Ans: \(I = \int {\frac{{{{\sin }^8}\,x – {{\cos }^8}\,x}}{{1 – 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x}}} dx\)
\( = \int {\frac{{\left( {{{\sin }^4}\,x + {{\cos }^4}\,x} \right)\left( {{{\sin }^4}\,x – {{\cos }^4}\,x} \right)}}{{{{\left( {{{\sin }^2}\,x + {{\cos }^2}\,x} \right)}^2} – 2\,{{\sin }^2}\,x\,{{\cos }^2}\,x}}} dx\)
\( = \int {\frac{{\left( {{{\sin }^4}\,x + {{\cos }^4}\,x} \right)\left( {{{\sin }^2}\,x + {{\cos }^2}\,x} \right)\left( {{{\sin }^2}\,x – {{\cos }^2}\,x} \right)}}{{{{\sin }^4}\,x + {{\cos }^4}\,x}}} dx\)
\( = \, – \int {\cos }\,2x\,dx\) \(\left[ {\because {{\cos }^2}\,x – {{\sin }^2}\,x = \cos \,2x} \right]\)
\( = – \frac{1}{2}\,\sin 2x + C\)
Hence, \(\int {\frac{{{{\sin }^3}\,x – {{\cos }^8}\,x}}{{1 – 2{{\sin }^2}\,x\,{{\cos }^2}\,x}}} dx = – \frac{1}{2}\sin 2x + C\)

Q.5. Evaluate : \(\int {{{\sin }^4}}\,x\,dx\)
Ans: Let \(I = \int {{{\sin }^4}}\,x\,dx\)
\( = \int {{{\left( {\frac{{1 – \cos 2x}}{2}} \right)}^2}} dx\left[ {\because {{\sin }^2}\theta = \frac{{1 – \cos 2\theta }}{2}} \right]\)
\( = \frac{1}{4}\int {\left( {1 – 2\cos 2x + {{\cos }^2}\,2x} \right)} dx\)
\( = \frac{1}{4}\int {\left( {1 – 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)} dx\)
\( = \frac{1}{8}\int {\left( {2 – 4\cos 2x + 1 + \cos 4x} \right)} dx\)
\( = \frac{1}{8}\int {\left( {3 – 4\cos 2x + \cos 4x} \right)} dx\)
\( = \frac{1}{8}\left\{ {3x – 2\sin 2x + \frac{{\sin 4x}}{4}} \right\} + C\)
Hence, \(\int {{{\sin }^4}\,xdx} = \frac{1}{8}\left\{ {3x – 2\sin 2x + \frac{{\sin 4x}}{4}} \right\} + C\)

Summary

To perform the integration of trigonometric functions, we must know its standard formulas, a few of which came from the differentiation of trigonometric functions. Recall trigonometric identities such as \(\cos \,2x = 2\,{\cos ^2}\,x – 1,\,\cos 2x = 1 – 2\,{\sin ^2}\,x,\,1 + {\tan ^2}\,x = {\sec ^2}\,x,\,{\cos ^2}\,x – {\sin ^2}\,x = \cos \,2x,\) etc. Remember that to find the value of \(\int {{{\sin }^m}}\,x\,{\cos ^n}\,x\,dx,\) where \(m,\) and \(n\) are non-negative integer, observe the exponents \(m,\) and \(n.\) For example, if \(m\) is odd, we substitute \(\cos \,x = t.\) Similarly, if \(n\) is odd, substitute \(\sin x = t.\) Also simplify the integrals having product of sine and cosine functions using the product to sum, difference formulae of trigonometric functions.

Frequently Asked Questions (FAQs)

Q.1. What is the integral of sine function?
Ans: The integration of \(\sin \,\theta \) is \( – \cos \,\theta ,\) and we mathematically, we write it as follows:
\(\int {\sin } \,\theta \,d\theta = – \cos \theta + C,\) where \(C\) is an arbitrary constant.

Q.2. What are the trigonometric integration formula?
Ans: Fundamental integration formulas of trigonometric functions are as follows:
1. \(\int {\sin } \,x\,dx = – \cos x + C\)
2. \(\int {\cos } \,x\,dx = \sin x + C\)
3. \(\int {{{\sec }^2}}\,x\,dx = \tan x + C\)
4. \(\int {{{\operatorname{cosec} }^2}}\,x\,dx = – \cot x + C\)
5. \(\int {\sec \,} x\tan x\,dx = \sec x + C\)
6. \(\int {\operatorname{cosec} } \,x\cot x\,dx = – \operatorname{cosec} x + C\)
7. \(\int {\cot } \,x\,dx = \log \left| {\sin x} \right| + C\)
8. \(\int {\sec } \,x\,dx = \log \left| {\sec x + \tan x} \right| + C\)
9. \(\int {\operatorname{cosec} \,} x\,dx = \log \left| {\operatorname{cosec} x – \cot x} \right| + C\)

Q.3. What are the 6 basic trigonometric functions?
Ans: In trigonometry, there are six functions of an angle that are often used. \({\text{Sine}}\left( {\sin } \right),\,{\text{cosine}}\left( {\cos } \right),\,{\text{tangent}}\left( {\tan } \right),\,{\text{cotangent}}\left( {\cot } \right),\,{\text{secant}}\left( {\sec } \right),\) and \({\text{cosecant}}\left( {{\text{cosec}}} \right)\) are their names and abbreviations.

Q.4. What are the integrals of the 6 trigonometric functions?
Ans: The integrals of \(6\) trigonometric functions are given as follows:
1. \(\int {\sin } \,x\,dx = – \cos x + C\)
2. \(\int {\cos } \,x\,dx = \sin x + C\)
3. \(\int {\tan } \,x\,dx = – \log \left| {\cos x} \right| + C\)
4. \(\int {\operatorname{cosec} } \,x\,dx = \log \left| {\operatorname{coses} x – \cot x} \right| + C\)
5. \(\int {\sec } \,x\,dx = \log \left| {\sec x + \tan x} \right| + C\)
6. \(\int {\cot } \,x\,dx = \log \left| {\sin x} \right| + C\)

Q.5. How do you integrate trigonometric functions with power?
Ans: To evaluate integrals of the form: \(\int {{{\sin }^m}}\,x\,dx,\int {{{\cos }^m}}\,x\,dx,\) where \(m \leqslant 4\), we express \({\sin ^m}\,x\) and \({\cos ^m}\,x\) in terms of sines and cosines of multiples of \(x\) by using the various trigonometrical identities. And, for large powers, we use the reduction formula’s given as follows:
1. \(\int {{{\sin }^n}}\,x\,dx = \frac{{ – \cos x \cdot {{\sin }^{n – 1}}\,x}}{n} + \frac{{n – 1}}{n}\int {{{\sin }^{n – 2}}}\,x\,dx\)
2. \(\int {{{\cos }^n}}\,x\,dx = \frac{{\sin x\,{{\cos }^{n – 1}}\,x}}{n} + \frac{{n – 1}}{n}\int {{{\cos }^{n – 2}}}\,x\,dx\)
3. \(\int {{{\tan }^n}}\,x\,dx = \frac{{{{\tan }^{n – 1}}\,x}}{{n – 1}} – \int {{{\tan }^{n – 2}}}\,x\,dx,n \geqslant 2\)
4. \(\int {{{\cot }^n}}\,x\,dx = – \frac{{{{\cot }^{n – 1}}\,x}}{{n – 1}} – {I_{n – 2}},n \geqslant 2\)
5. \(\int {{{\sec }^n}}\,x\,dx = \frac{{\tan x\,{{\sec }^{n – 2}}\,x}}{{n – 1}} + \frac{{n – 2}}{{n – 1}}{I_{n – 2}}\)

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