Integration in trigonometric functions helps us to find the antiderivatives of those functions by summing their infinitely small values over a given range. When applied to trigonometric functions, it reveals patterns and relationships that are vital in solving problems related to motion, waves, and periodic phenomena.
Mathematically, integration is the process of finding their antiderivatives by summing infinitesimally small values over a given range.
Formulas
The following is the list of integrals of the six basic trigonometric functions:
- ${\int \sin xdx=\cos x+C}$
- ${\int \cos xdx=-\sin x+C}$
- ${\int \sec ^{2}xdx=\tan x+C}$
- ${\int \text{cosec} ^{2}xdx=-\cot x+C}$
- ${\int \sec x\tan xdx=\sec x+C}$
- ${\int \text{cosec}\, x\cot xdx=-\text{cosec}\, x+C}$
By using trigonometric identities and substitution techniques, additional integrals can be derived:
- ${\int \tan xdx=\ln \left| \sec x\right| +C}$
- ${\int \cot xdx=\ln \left| \sin x\right| +C}$
- ${\int \sec xdx=\ln \left| \sec x+\tan x\right| +C}$
- ${\int \text{cosec}\, xdx=\ln \left| \text{cosec}\, x-\cot x\right| +C}$
Also, the powers of sine and cosine can be integrated by using trigonometric identities. For example,
- ${\int \sin ^{2}xdx=\dfrac{1}{2}\int \left( 1-\cos 2x\right) dx=\dfrac{x}{2}-\dfrac{\sin 2x}{4}+C}$
- ${\int \cos ^{2}xdx=\dfrac{1}{2}\int \left( 1+\cos 2x\right) dx=\dfrac{x}{2}+\dfrac{\sin 2x}{4}+C}$
When the integrand involves a product of sin mx and cos nx, the product-to-sum identities are used to simplify the expression. For example,
- ${\int \sin mx\cos nxdx}$ = ${\int \sin mx\cos nxdx=\dfrac{1}{2}\int \left[ \sin \left( m+n\right) x+\sin \left( m-n\right) x\right] dx}$ = ${-\dfrac{1}{2\left( m+n\right) }\cos \left( m+n\right) x-\dfrac{1}{2\left( m-n\right) }\cos \left( m-n\right) x+C}$
Here, C is an integral constant.
Now, let us look at several integrals of trigonometric functions and the techniques used to solve them through the following examples.
Using Trigonometric Identities
Trigonometric identities are used to simplify expressions before we integrate. These identities include:
- 2 sin A cos B = sin(A + B) + sin(A – B)
- 2 cos A cos B = cos(A – B) + cos(A + B)
- 2 sin A sin B = cos(A – B) – cos(A + B)
- sin2 A + cos2 A = 1
- sec2 A – tan2 A = 1
- cosec2 A – cot2 A = 1
- cos 2A = 2 cos2 A – 1
- cos 2A = 1 – 2 sin2 A
- sin 2A = 2 sin A cos A
Here are some examples showing how we integrate the given trigonometric functions using the identities.
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Evaluate ${\int \sin 3x\cos 2xdx}$
Solution:
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As we know,
The product-to-sum identity: sin mx cos nx = ${\dfrac{1}{2}\left[ \sin \left( m+n\right) x+\sin \left( m-n\right) x\right]}$
Here, sin 3x cos 2x = ${\dfrac{1}{2}\left[ \sin 5x+\sin x\right]}$
Given, ${\int \sin 3x\cos 2xdx}$
= ${\dfrac{1}{2}\int \left[ \sin 5x+\sin x\right] dx}$
= ${\dfrac{1}{2}\int \sin 5xdx+\dfrac{1}{2}\int \sin xdx}$
= ${\dfrac{1}{2}\left( -\dfrac{\cos 5x}{5}\right) +\dfrac{1}{2}\left( -\cos x\right) +C}$
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Evaluate: ${\int \dfrac{dx}{\sqrt{1-\sin ^{2}x}}}$
Solution:
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As we know, sin2 x + cos2 x = 1
⇒ 1 – sin2 x = cos2 x
⇒ ${\sqrt{1-\sin ^{2}x}}$ = cos x …..(i)
Given, ${\int \dfrac{dx}{\sqrt{1-\sin ^{2}x}}}$ …..(ii)
Substituting (i) in the expression (ii),
= ${\int \dfrac{dx}{\cos x}}$
= ${\int \sec xdx}$
= ${\ln \left| \sec x+\tan x\right| +C}$
Problem: Evaluating DEFINITE INTEGRALS
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Evaluate the definite integral: ${\int ^{\dfrac{\pi }{2}}_{0}\cos ^{2}xdx}$
Solution:
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As we know, a definite integral calculates the antiderivative value of a function over a specific interval [a, b]
Given, ${\int ^{\dfrac{\pi }{2}}_{0}\cos ^{2}xdx}$
Since cos2 x = ${\dfrac{1+\cos 2x}{2}}$
Here, ${\int ^{\dfrac{\pi }{2}}_{0}\cos ^{2}xdx}$
= ${\int ^{\dfrac{\pi }{2}}_{0}\dfrac{1+\cos 2x}{2}dx}$
= ${\dfrac{1}{2}\left[ \int ^{\dfrac{\pi }{2}}_{0}1dx+\int ^{\dfrac{\pi }{2}}_{0}\cos 2xdx\right]}$
= ${\dfrac{1}{2}\left[ x\right] _{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \dfrac{\sin 2x}{2}\right] _{0}^{\dfrac{\pi }{2}}}$
= ${\dfrac{\pi }{4}+\dfrac{1}{2}\left[ \dfrac{\sin \pi }{2}-\dfrac{\sin 0}{2}\right]}$
= ${\dfrac{\pi }{4}+0}$
= ${\dfrac{\pi }{4}}$
Methods for Solving Complex Integrals
To solve more complex integrals involving trigonometric functions, we use the substitution method, integration by parts, and appropriate trigonometric identities.
Using Substitution
This method simplifies integrals by replacing a complex trigonometric expression with a single variable.
Here are some examples showing how we integrate the given trigonometric functions using the substitution method.
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Solve: ${\int \sec ^{2}x\tan xdx}$
Solution:
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Here, we use the u-substitution method.
Let u = tan x ⇒ du = sec2 x dx …..(i)
Given, ${\int \sec ^{2}x\tan xdx}$ …..(ii)
Substituting (i) in the expression (ii),
= ${\int udu}$
Integrating,
= ${\dfrac{u^{2}}{2}+C}$
Since u = tan x
${\int \sec ^{2}x\tan xdx}$ = ${\dfrac{\tan ^{2}x}{2}+C}$
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Solve: ${\int \dfrac{\sin x}{1+\cos ^{2}x}dx}$
Solution:
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Let u = cos x ⇒ du = -sinx dx …..(i)
Given, ${\int \dfrac{\sin x}{1+\cos ^{2}x}dx}$ …..(ii)
Substituting (i) in the expression (ii),
= ${\int \dfrac{du}{1+u^{2}}}$
= ${-\arctan u+C}$
Since u = cos x
= ${-\arctan \left( \cos x\right) +C}$
Using By Parts
When a product of trigonometric and other functions is involved, use the formula:
${\int udv=uv-\int vdu}$
Here are a few examples showing how we integrate the given trigonometric functions using the by-parts method.
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Evaluate: ${\int x\sin xdx}$
Solution:
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As we know, using the integration by-parts formula,
${\int udv=uv-\int vdu}$ …..(i)
Let u = x and dv = sin x dx …..(ii)
⇒ du = dx and v = -cos x …..(iii)
Given, ${\int x\sin xdx}$ …..(iv)
Now, using (i), (ii), and (iii) in the expression (iv), we get
= ${-x\cos x+\int \cos xdx}$
= ${-x\cos x+\sin x+C}$
