Q91 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
The value of integral $\displaystyle\int \dfrac{dx}{\sin^2 x + \tan^2 x}$ is...
- A$\dfrac{-1}{2\tan x} + \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
- Correct answerB$\dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
- C$\dfrac{1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
- D$\dfrac{1}{2\tan x} + \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
Solution
Given integral is $I = \displaystyle\int \dfrac{dx}{\sin^2 x + \tan^2 x}$
Rewriting the denominator:
$\sin^2 x + \tan^2 x = \sin^2 x + \dfrac{\sin^2 x}{\cos^2 x} = \sin^2 x \left( \dfrac{\cos^2 x + 1}{\cos^2 x} \right) = \tan^2 x (\cos^2 x + 1)$
Substituting this back into the integral:
$I = \displaystyle\int \dfrac{dx}{\tan^2 x (\cos^2 x + 1)}$
Multiplying the numerator and the denominator by $\sec^2 x$:
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (\cos^2 x + 1) \sec^2 x} dx$
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (1 + \sec^2 x)} dx$
Using the identity $\sec^2 x = 1 + \tan^2 x$:
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (2 + \tan^2 x)} dx$
Let $\tan x = t$, then $\sec^2 x dx = dt$. The integral becomes:
$I = \displaystyle\int \dfrac{dt}{t^2 (t^2 + 2)}$
Using partial fractions:
$\dfrac{1}{t^2 (t^2 + 2)} = \dfrac{1}{2} \left( \dfrac{1}{t^2} - \dfrac{1}{t^2 + 2} \right)$
$I = \dfrac{1}{2} \displaystyle\int \left( \dfrac{1}{t^2} - \dfrac{1}{t^2 + (\sqrt{2})^2} \right) dt$
Integrating with respect to $t$:
$I = \dfrac{1}{2} \left( -\dfrac{1}{t} - \dfrac{1}{\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) \right) + c$
$I = \dfrac{-1}{2t} - \dfrac{1}{2\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) + c$
Substituting $t = \tan x$ back into the expression:
$I = \dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}} \tan^{-1} \left( \dfrac{\tan x}{\sqrt{2}} \right) + c$
Answer: $\dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
Rewriting the denominator:
$\sin^2 x + \tan^2 x = \sin^2 x + \dfrac{\sin^2 x}{\cos^2 x} = \sin^2 x \left( \dfrac{\cos^2 x + 1}{\cos^2 x} \right) = \tan^2 x (\cos^2 x + 1)$
Substituting this back into the integral:
$I = \displaystyle\int \dfrac{dx}{\tan^2 x (\cos^2 x + 1)}$
Multiplying the numerator and the denominator by $\sec^2 x$:
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (\cos^2 x + 1) \sec^2 x} dx$
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (1 + \sec^2 x)} dx$
Using the identity $\sec^2 x = 1 + \tan^2 x$:
$I = \displaystyle\int \dfrac{\sec^2 x}{\tan^2 x (2 + \tan^2 x)} dx$
Let $\tan x = t$, then $\sec^2 x dx = dt$. The integral becomes:
$I = \displaystyle\int \dfrac{dt}{t^2 (t^2 + 2)}$
Using partial fractions:
$\dfrac{1}{t^2 (t^2 + 2)} = \dfrac{1}{2} \left( \dfrac{1}{t^2} - \dfrac{1}{t^2 + 2} \right)$
$I = \dfrac{1}{2} \displaystyle\int \left( \dfrac{1}{t^2} - \dfrac{1}{t^2 + (\sqrt{2})^2} \right) dt$
Integrating with respect to $t$:
$I = \dfrac{1}{2} \left( -\dfrac{1}{t} - \dfrac{1}{\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) \right) + c$
$I = \dfrac{-1}{2t} - \dfrac{1}{2\sqrt{2}} \tan^{-1} \left( \dfrac{t}{\sqrt{2}} \right) + c$
Substituting $t = \tan x$ back into the expression:
$I = \dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}} \tan^{-1} \left( \dfrac{\tan x}{\sqrt{2}} \right) + c$
Answer: $\dfrac{-1}{2\tan x} - \dfrac{1}{2\sqrt{2}}\tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + c$
Hence, the correct answer is option: 'B'.