Trigonometric Integrals

Trigonometric Integrals

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y concerning x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point. These functions are also known as arcus functions, cyclometric functions, or anti-trigonometric functions. These functions are used to get an angle for a given trigonometric value. It refers to the change in the value of the trigonometric function at a certain rate.

(a) Integral of the form

1. $\int \frac{1}{a \cos ^2 x+b \sin ^2 x} d x$
2. $\int \frac{1}{a+b \sin ^2 x} d x$
3. $\int \frac{1}{a+b \cos ^2 x} d x$
4. $\int \frac{1}{a+b \sin ^2 x+c \cos ^2 x} d x$

Working Rule:

Step 1: Divide the numerator and denominator both by $\cos ^2 x$.
Step 2 : Put $\tan x=t, \sec ^2 x \mathrm{dx}=\mathrm{dt}$

This substitution will convert the trigonometric integral into an algebraic integral.

After employing these steps the integral will reduce to the form $\int \frac{f(t) d t}{A t^2+B t+C}$, where $f(t)$ is a polynomial in $t$.
This integral can be evaluated by methods we studied in previous concepts.

(b) Integral of the form

1. $\int \frac{1}{a \sin x+b \cos x} d x$
2. $\int \frac{1}{a+b \sin x} d x$
3. $\int \frac{1}{a+b \cos x} d x$
4. $\int \frac{1}{a \sin x+b \cos x+c} d x$

Working Rule:

Write sin x and cos x in terms of tan (x/2) and then substitute for tan (x/2) = t

i.e.

$\sin x=\frac{2 \tan x / 2}{1+\tan ^2 x / 2}$ and $\cos x=\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}$
replace, $\tan (\mathrm{x} / 2)$ with $t$
by performing these steps the integral reduces to the form
$\int \frac{1}{a t^2+b t+c} \mathrm{dt}$ which can be solved by method we studied in previous concepts.

(c) Integrals of the form

1. $\int \frac{p \cos x+q \sin x+r}{a \cos x+b \sin x+c} d x$
2. $\int \frac{p \cos x+q \sin x}{a \cos x+b \sin x} d x$

Working Rule:

Express numerator as $\lambda($ denominator $)+\mu($ differentiation of denominator $)+\gamma$

$\Rightarrow(\mathrm{p} \cos \mathrm{x}+\mathrm{q} \sin \mathrm{x}+\mathrm{r})=\lambda(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}+\mathrm{c})+\mu(-\mathrm{a} \sin \mathrm{x}+\mathrm{b} \cos x)+\gamma$

where $\lambda, \mu$, and \gamma are constants to be determined by comparing the coefficients of $\sin x, \cos \mathrm{x}$, and constant terms on both sides.

$\begin{aligned}
\int \frac{p \cos \mathrm{x}+\mathrm{q} \sin \mathrm{x}+\mathrm{r}}{a \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}+\mathrm{c}} \mathrm{dx}= & \int \frac{\lambda(\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}+\mathrm{c})+\mu(-\mathrm{a} \sin \mathrm{x}+\mathrm{b} \cos \mathrm{x})+\gamma}{\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}+\mathrm{c}} \mathrm{dx} \\
= & \lambda \int \frac{a \cos x+b \sin x+c}{a \cos x+b \sin x+c} d x+\mu \int \frac{-a \sin x+b \cos x}{a \cos x+b \sin x+c} d x \\
& \quad+\int \frac{\mu}{\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}+\mathrm{c}} \mathrm{dx} \\
= & \lambda x+\mu \ln |a \cos x+b \sin x+c|+\int \frac{\mu}{a \cos x+b \sin x+c} d x
\end{aligned}$

Recommended Video Based on Trigonometric Integrals

Solved Questions Based on Trigonometric Integrals

Example 1: Integrate $\int \frac{d x}{\sin ^2 x+2 \sin x \cos x}$

1) $\ln \frac{t+1}{t-1}+c$
2) $\ln \frac{t+2}{t}+c$
3) $\frac{1}{2} \ln \frac{t}{(t+2)}+C$
4) none of these

Solution
Divide by $\cos ^2 x$ in each case and pull $t=\tan x, d t=\sec ^2 x d x$

$\int \frac{d x \sec ^2 x}{\tan ^2 x+2 \tan x}=\int \frac{d t}{(t+1)^2-1}=\frac{1}{2} \ln \frac{t}{(t+2)}+C$

Hence, the answer is the option 3.
Example 2: $\int \frac{d x}{1+\sin x}$
1) $\frac{1}{1+\tan \left(\frac{x}{2}\right)}+C$

2) $-\frac{2}{\tan \left(\frac{x}{2}\right)+1}+C$
3) $\frac{1}{1+\cot \left(\frac{x}{2}\right)}+C$
4) $-\frac{1}{1+\cot \left(\frac{x}{2}\right)}+C$

$\begin{aligned}
& I=\int \frac{1}{\sin (x)+1} \mathrm{~d} x \\
& =\int \frac{\sec ^2\left(\frac{x}{2}\right)}{\left(\tan \left(\frac{x}{2}\right)+1\right)^2} \mathrm{~d} x \\
& \text { Put } u=\tan \left(\frac{x}{2}\right)+1 \Rightarrow \mathrm{d} x=\frac{2}{\sec ^2\left(\frac{x}{2}\right)} \mathrm{d} u \\
& I=2 \int \frac{1}{u^2} d u \\
& =-\frac{2}{u} \\
& =-\frac{2}{\tan \left(\frac{x}{2}\right)+1}+C
\end{aligned}$

Hence, the answer is the option (2).

Example 3: $\int \frac{d x}{3+4 \cos ^2 x}$ equals
1) $\frac{\sqrt{3}}{\sqrt{7}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$
2) $\frac{1}{\sqrt{21}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$

$\sqrt{3} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C$

4) None of these

Solution
Divide numerator and denominator by $\cos ^2 x$

$\begin{aligned}
& \int \frac{\sec ^2 x d x}{3 \sec ^2 x+4} \\
& =\int \frac{\sec ^2 x d x}{3 \tan ^2 x+3+4} \\
& =\int \frac{\sec ^2 x d x}{7+3 \tan ^2 x}
\end{aligned}$

Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$

$\begin{aligned}
& \int \frac{d t}{7+3 t^2} \\
& =\frac{1}{3} \int \frac{d t}{t^2+\left(\sqrt{\frac{7}{3}}\right)^2} \\
& =\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{7}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C \\
& =\frac{1}{\sqrt{21}} \tan ^{-1}\left(\sqrt{\frac{3}{7}} \tan x\right)+C
\end{aligned}$

Hence, the answer is the option 2.

Example 4: Integrate $\int \frac{d x}{\sin ^2 x+2 \cos ^2 x+2 \sin x \cos x}$
1) $x+c$
2) $\frac{1}{2} \tan ^{-1}(1+x)+C$
3) $\tan ^{-1}(1+\tan x)+C$

4) None of these

Solution

Divide by $\cos^{2} x$

$\begin{aligned}
& \int \frac{\sec ^2 x \cdot d x}{\tan ^2 x+2+2 \tan x} \\
& I=\int \frac{\sec ^2 x d x}{(\tan x+1)^2+1}
\end{aligned}$

Put $\tan (x)=t$

$\sec ^2 x d x=d t$

$\begin{aligned}
& I=\int \frac{d t}{(t+1)^2+1} \\
& =\tan ^{-1}(t+1)+C \\
& =\tan ^{-1}(\tan x+1)+C
\end{aligned}$

Hence, the answer is the option 3.

Example 5: $\int \frac{d x}{1+2 \cos x}$
1) $\frac{1}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \frac{x}{2}}{\sqrt{3}-\tan \frac{x}{2}}\right|+c$
2) $-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{\tan \left(\frac{x}{2}\right)-\sqrt{3}}{\tan \left(\frac{x}{2}\right)+\sqrt{3}}\right|+c$
3) $\frac{2}{\sqrt{3}} \ln \left|\frac{\sqrt{3}+\tan \frac{x}{2}}{\sqrt{3}-\tan \frac{x}{2}}\right|+c$
4) None of these

Solution

$\begin{aligned} & \int \frac{d x}{1+2 \cos x} \\ & \int \frac{d x}{1+\frac{2-2 \tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{2}{2}}} \\ & =\int \frac{\sec ^2\left(\frac{x}{2}\right) \cdot d x}{3-\tan ^2\left(\frac{x}{2}\right)} \\ & \text { Put } u=\tan \left(\frac{x}{2}\right) \Rightarrow \frac{1}{2} \sec ^2\left(\frac{x}{2}\right) \cdot d x=d u \\ & =-2 \int \frac{1}{u^2-3} d u \\ & =-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{u-\sqrt{3}}{u+\sqrt{3}}\right|+c \\ & =-\frac{1}{\sqrt{3}} \cdot \ln \left|\frac{\tan \left(\frac{x}{2}\right)-\sqrt{3}}{\tan \left(\frac{x}{2}\right)+\sqrt{3}}\right|+c\end{aligned}$Hence, the answer is the option (2).

Summary

Trigonometric integral is an important concept used in practical and traditional approaches in mathematics. Trigonometric integrals are integrals that involve trigonometric functions such as sine, cosine, tangent, and their inverses.