Question

What is $${\int }_0^1{\displaystyle \frac{{\tan }^{-1}x}{1+x^2}}dx$$ equal to?
A.$${\displaystyle \frac{{\pi }^2}{8}}$$
B.$${\displaystyle \frac{{\pi }^2}{32}}$$
C.$${\displaystyle \frac{\pi }{4}}$$
D.$${\displaystyle \frac{\pi }{2}}$$

Answer 1

Hint: Here in this question, we have to find the integrated value of a given trigonometric function. This can be solved by using a substitution method and later integrated by using the standard trigonometric formula of integration. Since they have mentioned the limit points. It is a definite integral on applying the limits we get the required solution.

Complete step by step solution:
In integration we have two different kinds. One is definite integral and another one is indefinite integral. In definite integral the limits points are mentioned. In indefinite integral the limit points are not mentioned.
Here this question belongs to the definite integral where the limits points are mentioned.
Now consider the given function
$${\int }_0^1{\displaystyle \frac{{\tan }^{-1}x}{1+x^2}}dx$$--------(1)
Here the lower limit is 0 and the upper limit is 1
We integrate the above function by using a substitution method
Let take, $$t={\tan }^{-1}x$$
Differentiate with respect to x, then
$$\Rightarrow {\displaystyle \frac{dt}{dx}}={\displaystyle \frac{1}{1+x^2}}$$
Or
$$\Rightarrow dt={\displaystyle \frac{1}{1+x^2}}dx$$
Substitute $$t$$ and $$dt$$ in equation (1), we have
 $$\Rightarrow {\int }_0^{1}tdt$$
Now integrate using a formula $${\int }_a^b{x}^n={{\displaystyle \frac{x^{n+1}}{n+1}}|}_a^b$$, then we gave
$$\Rightarrow {{\displaystyle \frac{t^2}{2}}|}_0^1$$
Put, $$t={\tan }^{-1}x$$, then
 $$\Rightarrow {{\displaystyle \frac{1}{2}}{\left({\tan }^{-1}x\right)}^2|}_0^1$$
On applying a limit, we have
$$\Rightarrow {\displaystyle \frac{1}{2}}\left[{\left({\tan }^{-1}1\right)}^2-{\left({\tan }^{-1}0\right)}^2\right]$$
As we know the value of $${\tan }^{-1}1={\displaystyle \frac{\pi }{4}}$$ and $${\tan }^{-1}0=0$$, then on substituting the values we get
$$\Rightarrow {\displaystyle \frac{1}{2}}\left[{\left({\displaystyle \frac{\pi }{4}}\right)}^2-{\left(0\right)}^2\right]$$
On simplification, we get
$$\Rightarrow {\displaystyle \frac{1}{2}}\left[{\left({\displaystyle \frac{\pi }{4}}\right)}^2\right]$$
$$\Rightarrow {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{{\pi }^2}{16}}\right]$$
$$\Rightarrow {\displaystyle \frac{{\pi }^2}{32}}$$
Hence, we have integrated the given function and applied the limit points and obtained an answer.
$$\therefore$$ The value of $${\int }_0^1{\displaystyle \frac{{\tan }^{-1}x}{1+x^2}}dx={\displaystyle \frac{{\pi }^2}{32}}$$. So, option (B) is correct.

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.