Antiderivatives

# Standard Integrals (III)

Here, we will discuss about the various techniques that may be applied for evaluating standard integrals of some special trigonometric functions. [Antiderivatives]

## Standard Integrals of Some Trigonometrical Functions

### $\int\operatorname{cosec}xdx=\log\left(\tan\frac{x}{2}\right)+C$

$\int\operatorname{cosec}xdx=\int\frac{1}{\sin x}dx=\int\frac{dx}{2\sin\frac{x}{2}\cos\frac{x}{2}}$ $=\frac{1}{2}\int\frac{1}{\sin\frac{x}{2}\cos\frac{x}{2}}×\frac{\sec^2\frac{x}{2}}{\sec^2\frac{x}{2}}dx$ $=\frac{1}{2}\int\frac{\sec^2\frac{x}{2}}{\tan\frac{x}{2}}dx=\log\left(\tan\frac{x}{2}\right)+C$

### $\int\sec xdx=\log\left(\tan\left(\frac{π}{4}+\frac{x}{2}\right)\right)+C$

$\int\sec xdx=\int\frac{dx}{\cos x}=\int\frac{dx}{\sin\left(\frac{π}{2}+x\right)}$ $=\frac{1}{2}\int\frac{dx}{\sin\left(\frac{π}{4}+\frac{x}{2}\right)\cos\left(\frac{π}{4}+\frac{x}{2}\right)}$ $=\frac{1}{2}\int\frac{\sec^2\left(\frac{π}{4}+\frac{x}{2}\right)}{\tan\left(\frac{π}{4}+\frac{x}{2}\right)}dx$ $=\log\left(\tan\left(\frac{π}{4}+\frac{x}{2}\right)\right)+C$

### $\operatorname{cosech}xdx=\log\left(\tanh\frac{x}{2}\right)+C$

$\text{Let }I=\operatorname{cosech}xdx$ $=\int\frac{1}{\sinh x}dx=\int\frac{2}{e^x-e^{-x}}dx=\int\frac{2e^x}{e^{2x}-1}$ Put $e^x=y$, then, $e^xdx=dy$. $\therefore I=2\int\frac{dy}{y^2-1}=\log\frac{y-1}{y+1}+C$ $=\log\frac{e^x-1}{e^x+1}+C=\log\left(\tanh\frac{x}{2}\right)+C$

### $\operatorname{sech}xdx=2\tan^{-1}\left(\tanh\frac{x}{2}\right)+C$

$\text{Let }I=\int\operatorname{sech}xdx=\int\frac{1}{\cosh x}dx$ $=\int\frac{dx}{\cosh^2\frac{x}{2}+\sinh^2\frac{x}{2}}=\int\frac{\operatorname{sech}^2\frac{x}{2}}{1+\tanh^2\frac{x}{2}}dx$ Put $\tanh\frac{x}{2}=y$, then, $\operatorname{sech}^2\frac{x}{2}dx=2dy$. $\therefore I=2\int\frac{dy}{1+y^2}=2\tan^{-1}y+C$ $=2\tan^{-1}\left(\tanh\frac{x}{2}\right)+C$

### $\int\frac{dx}{a+b\cos x}$

$\int\frac{dx}{a+b\cos x}$ $=\int\frac{dx}{a\left(\cos^2\frac{x}{2}+\sin^2\frac{x}{2}\right)+b\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)}$ $=\int\frac{1}{(a+b)\cos^2\frac{x}{2}+(a-b)\sin^2\frac{x}{2}}×\frac{\sec^2\frac{x}{2}}{\sec^2\frac{x}{2}}dx$ $=\int\frac{\sec^\frac{x}{2}}{(a+b)+(a-b)\tan^2\frac{x}{2}}dx$ Case I. $a>b$

Put $\sqrt{a-b}.\tan\frac{x}{2}=z$, then $\frac{1}{2}\sqrt{a-b}\sec^2\frac{x}{2}dx=dz$. Then the given integral transforms into $\frac{2}{\sqrt{a-b}}\int\frac{dz}{(a+b)^2+z^2}$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\frac{z}{\sqrt{a+b}}+C$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)+C$ Case II. $a<b$

The given integral can be written as $\int\frac{\sec^2\frac{x}{2}}{(a+b)-(b-a)\tan^2\frac{x}{2}}dx\text{ __(1)}$ Put $\sqrt{b-a}\tan\frac{x}{2}=z$, then, $\frac{1}{2}\sqrt{b-a}\sec^2\frac{x}{2}dx=dz$ Then the integral $\text{(1)}$ becomes $\frac{2}{\sqrt{b-a}}\int\frac{dz}{(a+b)-z^2}$ $=\frac{2}{\sqrt{b-a}}.\frac{1}{2\sqrt{b+a}}\log\frac{\sqrt{b+a}+z}{\sqrt{b+a}-z}+C$ $=\frac{1}{\sqrt{b^2-a^2}}\log\left(\frac{\sqrt{b+a}+\sqrt{b-a}\tan\frac{x}{2}}{\sqrt{b+a}-\sqrt{b-a}\tan\frac{x}{2}}\right)+C$

### $\int\frac{dx}{a+b\sin x}$

$\text{Let }I=\int\frac{dx}{a+b\sin x}$ $=\int\frac{dx}{a\left(\cos^2\frac{x}{2}+\sin^2\frac{x}{2}\right)+2b\sin\frac{x}{2}\cos\frac{x}{2}}$ $=\int\frac{1}{a\cos^2\frac{x}{2}+a\sin^2\frac{x}{2}+2b\sin\frac{x}{2}\cos\frac{x}{2}}×\frac{\sec^2\frac{x}{2}}{\sec^2\frac{x}{2}}dx$ $=\int\frac{\sec^2\frac{x}{2}}{a+a\tan^2\frac{x}{2}+2b\tan\frac{x}{2}}dx$ Put $\tan\frac{x}{2}=y$, then, $\sec^2\frac{x}{2}=2dy$. $\therefore I=\int\frac{2dy}{a+ay^2+2by}=2\int\frac{dy}{ay^2+2by+a}$ $=2\int\frac{dy}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)^2-\left(\frac{b}{\sqrt{a}}\right)^2+a}$ $=2\int\frac{dy}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)^2-\left(\frac{b^2}{a}-a\right)}$ $=2\int\frac{dy}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)^2-\frac{(b^2-a^2)}{a}}$

Case I: $b>a$, $I=\int\frac{dy}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)^2-\frac{(b^2-a^2)}{a}}$ Put $\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)=z$, then, $\sqrt{a}dy=dz$ $\Rightarrow dy=\frac{dz}{\sqrt{a}}$ $\therefore I=\frac{2}{\sqrt{a}}\int\frac{dz}{z^2-\left(\frac{\sqrt{b^2-a^2}}{\sqrt{a}}\right)^2}$ $=\frac{2}{\sqrt{a}}×\frac{1}{2\frac{\sqrt{b^2-a^2}}{\sqrt{a}}}\log\frac{z-\frac{\sqrt{b^2-a^2}}{\sqrt{a}}}{z+\frac{\sqrt{b^2-a^2}}{\sqrt{a}}}+C$ $=\frac{1}{\sqrt{b^2-a^2}}\log\frac{z\sqrt{a}-\sqrt{b^2-a^2}}{z\sqrt{a}+\sqrt{b^2-a^2}}+C$ $=\frac{1}{\sqrt{b^2-a^2}}\log\frac{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)\sqrt{a}-\sqrt{b^2-a^2}}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)\sqrt{a}+\sqrt{b^2-a^2}}+C$ $=\frac{1}{\sqrt{b^2-a^2}}\log\frac{ay+b-\sqrt{b^2-a^2}}{ay+b+\sqrt{b^2-a^2}}+C$ $=\frac{1}{\sqrt{b^2-a^2}}\log\frac{a\tan\frac{x}{2}+b-\sqrt{b^2-a^2}}{a\tan\frac{x}{2}+b+\sqrt{b^2-a^2}}+C$

Case II: $a>b$, $I=2\int\frac{dy}{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)^2+\frac{(a^2-b^2)}{a}}$ Put $\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)=z$, then, $\sqrt{a}dy=dz$ $\Rightarrow dy=\frac{dz}{\sqrt{a}}$ $\therefore I=\frac{2}{\sqrt{a}}\int\frac{dz}{z^2+\left(\frac{\sqrt{a^2-b^2}}{\sqrt{a}}\right)^2}$ $=\frac{2}{\sqrt{a}}×\frac{1}{\frac{\sqrt{a^2-b^2}}{\sqrt{a}}}\tan^{-1}\frac{z}{\frac{\sqrt{a^2-b^2}}{\sqrt{a}}}+C$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\frac{z\sqrt{a}}{\sqrt{a^2-b^2}}+C$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\frac{\left(y\sqrt{a}+\frac{b}{\sqrt{a}}\right)\sqrt{a}}{\sqrt{a^2-b^2}}+C$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\frac{ay+b}{\sqrt{a^2-b^2}}+C$ $=\frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\left(\frac{a\tan\frac{x}{2}+b}{\sqrt{a^2-b^2}}\right)+C$

## Standard Integral of the Form (III)

An integral of the form $\int\frac{dx}{a\sin x+b\cos x}$ can be obtained in the following way.

Put $a=r\cos\theta$ and $b=r\sin\theta$, so that $r=\sqrt{a^2+b^2}$ and $\theta=\tan^{-1}\frac{b}{a}$. $\therefore\int\frac{dx}{a\sin x+b\cos x}=\frac{1}{r}\int\frac{dx}{\cos\theta\sin x+\sin\theta\cos x}$ $=\frac{1}{r}\int\frac{dx}{\sin(x+\theta)}=\frac{1}{r}\int\operatorname{cosec}(x+\theta)dx$ $=\frac{1}{r}\log\left(\tan\frac{1}{2}(x+\theta)\right)+C$ $=\frac{1}{\sqrt{a^2+b^2}}\log\left(\tan\frac{1}{2}\left(x+\tan^{-1}\frac{b}{a}\right)\right)+C$

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