Evaluate: {² x}{{16+² x}} dx

Correct Answer is : log⁡∣tan⁡x+16+tan⁡2x∣+C ≤ft| x + {16+² x} | + Clogtanx+16+tan2x+C

Correct Answer is : log⁡∣tan⁡x+16+tan⁡2x∣+C ≤ft| x + {16+² x} | + Clogtanx+16+tan2x+C

Test Index

UPSC NDA Math Model Paper 5

Question : 24 of 120

Marks: +1, -0

Evaluate:

\int \frac{\sec^2 x}{\sqrt{16+\tan^2 x}} dx

Solution:

Here we have to find the value of

\int \frac{\sec^2 x}{\sqrt{16+\tan^2 x}} dx

Let

\tan x = t

then

\sec^2 x dx = dt

\Rightarrow \int \frac{\sec^2 x}{\sqrt{16+\tan^2 x}} dx = \int \frac{dt}{\sqrt{16+t^2}}

Now by comparing

\int \frac{dt}{\sqrt{16+t^2}}

with

\int \frac{dx}{\sqrt{x^2 + a^2}}

we get,

a = 4

As we know that,

\int \frac{dx}{\sqrt{x^2 + a^2}} = \log x + \left| \sqrt{x^2 + a^2} \right| + C

where

C

is a constant

\Rightarrow \int \frac{dt}{\sqrt{16+t^2}} = \log \left| t + \sqrt{t^2+16} \right| + C

Now by substituting

\tan x = t

in the above equation we get,

\int \frac{\sec^2 x}{\sqrt{16+\tan^2 x}} dx

= \log \left| \tan x + \sqrt{16+\tan^2 x} \right| + C

where

C

is a constant

Go to Question:

Prev Question

Next Question