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Evaluate:$\Large\int \limits_{\normalsize 0}^{\normalsize \pi} $$\large\frac{x\; \tan x}{\sec x\; \text{cosec} x}$


$(B) \large\frac{\pi^2}{4}$

$(C) \large\frac{3\pi^2}{2}$

$ (D) \large\frac{\pi^2}{2}$

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  • $(i)\; \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $(ii)\; \tan x=\large\frac{\sin x}{\cos x}$
  • $(iii)\; \sec x=\large\frac{1}{\cos x}$
  • $(iv)\; cosec x=\large\frac{1}{\sin x}$
  • $(v)\; \bigg[\large\frac{1-\cos 2 x}{2}\bigg]=$$\sin ^2 x$
Consider $\large\frac{x \;\tan x}{\sec x \;cosec x}$
This can be written as $\Large \frac{x.\frac{\sin x}{\cos x}}{\frac{1}{\cos x}.\frac{1}{\sin x}}$
$=\large\frac{x\; \sin x}{\cos x} $$\times \sin x \cos x$
$=x \sin ^2 x$
But $ \sin ^2 x =\large\frac{1-\cos 2x}{2}$
Therefore $ x \sin ^2 x=\large\frac{x(1-\cos 2x)}{2}$
$=\large\frac{x}{2}- \frac{x \cos 2x}{2}$
Hence $I= \int \limits_0^ \pi (\large\frac{x}{2} -\frac{x \cos 2x}{2})$$ dx=\int \limits_0^{\pi} \large\frac{x}{2}-\frac{1}{2} \int \limits_0^{\pi} $$x \cos 2x dx$-----(1)
Consider $ \large\frac{1}{2} $$\int x \cos 2x dx$
This is of the form $\int udv$
we know $\int udv=uv-\int vdu$
Here Let $u=x$ on differentiating with respect to we get, $du=dx$
Let $ \cos 2x dx=dv$ on integrating we get $\large\frac{1}{2} $$\sin 2x =v$
Therefore substituting for u,v,du and dv we get,
$\large\frac{1}{2} $$\int x \cos 2x dx$$=\large\frac{1}{2} $$\bigg\{\bigg[ x (\large\frac{1}{2} $$\sin 2x)-\frac{1}{2}\int \sin 2x .dx\bigg]\bigg\}$
$=\large\frac{x \sin 2x}{4} -\frac{1}{4} $$\int \sin 2xdx$
On integrating
$=\large\frac{x \sin 2x}{4} -\frac{1}{4} (\frac{-\cos 2x}{2})$
$=\large\frac{x \sin 2x}{4} +\frac{\cos 2x}{8}$
Now substituting in equ (1)
we know $\int \large\frac{x}{2}=\frac{x^2}{4}$
Therefore $ \int \limits_0^{\pi} \large\frac{x}{2}$$dx-\int \limits_0^{\pi} \large\frac{x \cos 2x}{2} $$dx= \bigg[ \large\frac{x^2}{4}\bigg]_0^{\pi}-\bigg[ \frac{x \sin 2x}{4}+\frac{\cos 2x}{8}\bigg]_0^{\pi}$
Now applying the limits,
$I=\bigg(\large\frac{\pi ^2}{4}-\normalsize 0 \bigg)-\bigg[ \frac{\pi \sin 2 \pi}{4}-\normalsize 0-\frac{\cos 2 \pi}{8} +\frac{\cos 0}{8}\bigg]$
we know $\sin 2\pi =0, \cos 2 \pi =\cos 0=1$
Therefore $I=\large\frac {\pi^2}{4}-\frac{1}{8}+\frac{1}{8}$
answered Apr 25, 2013 by meena.p
edited Mar 25, 2014 by balaji.thirumalai
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