Q&A - Ask Doubts and Get Answers
Q

Evaluate the definite integrals in Exercises 25 to 33. 31

Evaluate the definite integrals in Exercises 25 to 33.

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

Answers (1)
Views

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

Exams
Articles
Questions