Question

The value of integral \(\int \frac{dx}{sin^2x+tan^2x}\) is...

• A. \(\frac{-1}{2tanx}+\frac{1}{2\sqrt{2}}tan^{-1}(\frac{tanx}{\sqrt{2}})+c\)
• B. \(\frac{-1}{2tanx}-\frac{1}{2\sqrt{2}}tan^{-1}(\frac{tanx}{\sqrt{2}})+c\)
• C. \(\frac{1}{2tanx}-\frac{1}{2\sqrt{2}}tan^{-1}(\frac{tanx}{\sqrt{2}})+c\)
• D. \(\frac{1}{2tanx}+\frac{1}{2\sqrt{2}}tan^{-1}(\frac{tanx}{\sqrt{2}})+c\)

Hint:

Convert to tan x using t = tan x, then use partial fractions.

Answer 1

The Correct Option is B

Step 1: Rewrite in Terms of tan x:
Let \(t=\tan x\), so \(dx=\dfrac{dt}{1+t^2}\), \(\sin^2x=\dfrac{t^2}{1+t^2}\). Then
\[ \sin^2x+\tan^2x=\frac{t^2}{1+t^2}+t^2=\frac{t^2(2+t^2)}{1+t^2} \]

Step 2: Form the Integral:
\[ I=\int\frac{1+t^2}{t^2(2+t^2)}\cdot\frac{dt}{1+t^2}=\int\frac{dt}{t^2(t^2+2)} \]

Step 3: Partial Fractions:
\(\dfrac1{t^2(t^2+2)}=\dfrac12\left(\dfrac1{t^2}-\dfrac1{t^2+2}\right)\). So
\[ I=\frac12\left[-\frac1t-\frac1{\sqrt2}\tan^{-1}\frac t{\sqrt2}\right]+c \]

Step 4: Back-Substitute:
\[ I=-\frac1{2\tan x}-\frac1{2\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+c \]
Option (B) matches. Options (A) and (D) have a plus sign in front of the arctan term, and (C) has the wrong sign for \(1/(2\tan x)\).

Final Answer:
The integral is option (B). \[ \boxed{\text{(B)}} \]