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Evaluate the definite integrals in Exercises 25 to 33.

Q31. $\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx$

Answers (1)

best_answer

Let I =

$\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

$=\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

$\\t=sinx \\dt=cosxdx$

Now the important step here is to change the limit of the integration as we are changing the variable.so,

$\\when\:x=0,t=sin0=0 \\when\:x=\frac{\pi}{2},t=sin\frac{\pi}{2}=1$

So our function becomes,

$I=2\int_{0}^{1}(tan^{-1}t)tdt$

Now, let's integrate this by using integration by parts method,

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{1+t^2}\cdot\frac{t^2}{2}dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{t^2}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{(1+t^2)-1}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\left ( 1-\frac{1}{1+t^2} \right )\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t-tan^{-1}t) \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t)+\frac{1}{2}tan^{-1}t) \right ]_0^1$

$I=2\left [ \frac{1}{2} \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}(1)\cdot(1^2+1)-1 \right )\right ]-\left [ \left (tan^{-1}(0)\cdot(0^2+1)-0 \right )\right ]$ $I=2tan^{-1}1-1=2\times \frac{\pi}{4}-1$

$I=\frac{\pi}{2}-1$

Posted by

Pankaj Sanodiya

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