Question
Evaluate the following integrals:$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$

Answer

Let $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ....(\text{i})$$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec(\pi-\text{x})\text{ cosec}(\pi-\text{x})}\text{ dx}$ $\Bigg[\text{Using}\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0\frac{-(\pi-\text{x})\tan\text{x}}{-\sec\text{x}\text{ cosec}\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^2\text{x dx}$
$=\pi\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\Big[\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The contents of urns I, II, III are as follows:
Urn I : 1 white, 2 black and 3 red balls
Urn II : 2 white, 1 black and 1 red balls
Urn III : 4 white, 5 black and 3 red balls.
One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I, II, III?
Show that the line $\frac{\text{x}+3}{-3}=\frac{\text{y}-1}{1}=\frac{\text{z}-5}{5}$ and $\frac{\text{x}+1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-5}{5}$ are coplanar. Hence, find the equation of the plane containing these lines.
Show that the points (3, 4), (-5, 16) and (5, 1) are collinear.
Evaluate the following definite integrals:$\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Find the vector equation of a line passing through the point with position vector $\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$ and parallel to the line joining the points with position vectors $\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$ Also, find the cartesian equivalent of this equation.
Solve the following differential equations:$(\text{xy}^2+2\text{x})\text{dx}+(\text{x}^2\text{y+2y})\text{dy}=0$
Find $X$, if $A X=B$ where $A=\left[\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$.
If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$