{ x}{² x ( x + x)} \, dx is equal to

Correct Answer is : log⁡∣1+tan⁡xtan⁡x∣−cot⁡x+C ≤ft| 1+ x/ x | - x + Clogtanx1+tanx−cotx+C

Correct Answer is : log⁡∣1+tan⁡xtan⁡x∣−cot⁡x+C ≤ft| 1+ x/ x | - x + Clogtanx1+tanx−cotx+C

Test Index

Manipal Entrance Test 2014 Math Paper

Question : 44 of 80

Marks: +1, -0

\int \frac{\cos x}{\sin^{2} x (\sin x + \cos x)} \, dx

is equal to

Solution:

Let,

\int \frac{\cos x}{\sin^{2} x (\sin x + \cos x)} \, dx

On dividing numerator and denominator by

\cos^{3} x

, we get

I = \int \frac{\sec^{2} x}{\tan^{2} x (1+\tan x)} \, dx

Put

\tan x = t \Rightarrow \sec^{2} x \, dx = dt

∴

I = \int \frac{dt}{t^{2}(1+t)}

= \int \left[ -\frac{1}{t} + \frac{1}{t^{2}} + \frac{1}{1+t} \right] dt

(using partial fraction)

⇒

I = -\int \frac{1}{t} \, dt + \int t^{-2} \, dt + \int \frac{1}{1+t} \, dt

= -\log |t| - \frac{1}{t} + \log |1+t| + C

= -\frac{1}{\tan x} + \log \left| \frac{1+\tan x}{\tan x} \right| + C

= -\cot x + \log \left| \frac{1+\tan x}{\tan x} \right| + C

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