Correct Answer is : 14tan⁡−1(12tan⁡x)+c1/4 ^{-1}≤ft(1/2 x)+c41tan−1(21tanx)+c, where c is the constant of integration.

1/2 ^{-1}( x)+c, where c is the constant of integration.

1/2 ^{-1}≤ft( x/2)+c, where c is the constant of integration.

1/4 ^{-1}≤ft(1/2 x)+c, where c is the constant of integration.

1/4 ^{-1}( x)+c, where c is the constant of integration.

Correct Answer is : 14tan⁡−1(12tan⁡x)+c1/4 ^{-1}≤ft(1/2 x)+c41tan−1(21tanx)+c, where c is the constant of integration.

Test Index

MHTCET 25 Apr 2025 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

\int \frac{d x}{3 \cos 2 x+5}

equals

$\frac{1}{2} \tan ^{-1}(\tan x)+c$ , where c is the constant of integration.
$\frac{1}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c$ , where c is the constant of integration.
$\frac{1}{4} \tan ^{-1}\left(\frac{1}{2} \tan x\right)+c$ , where c is the constant of integration.
$\frac{1}{4} \tan ^{-1}(\tan x)+c$ , where c is the constant of integration.

Solution:

Concept:

Use the tangent half-angle substitution to convert the integrand into a standard inverse tangent form.

Explanation:

Let

I = \int \frac{dx}{3\cos 2x + 5}

Put

\tan x = t

Then

x = \tan^{-1} t

, so

dx = \frac{dt}{1+t^2}

Also,

\cos 2x = \frac{1-t^2}{1+t^2}

Substituting these into the integral gives

I = \int \frac{\frac{dt}{1+t^2}}{3\left(\frac{1-t^2}{1+t^2}\right)+5}.

Simplify the denominator:

3\left(\frac{1-t^2}{1+t^2}\right)+5 = \frac{3-3t^2+5+5t^2}{1+t^2} = \frac{8+2t^2}{1+t^2}.

Therefore,

I = \int \frac{dt}{8+2t^2} = \frac{1}{2}\int \frac{dt}{4+t^2}.

Using

\int \frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+c

with

a=2

and

u=t

, we get

I = \frac{1}{2} \cdot \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)+c = \frac{1}{4}\tan^{-1}\left(\frac{t}{2}\right)+c.

Substitute back

t = \tan x

I = \frac{1}{4}\tan^{-1}\left(\frac{\tan x}{2}\right)+c.

This matches option C.

Answer:

\frac{1}{4}\tan^{-1}\left(\frac{1}{2}\tan x\right)+c

, i.e. option C.

Go to Question:

Prev Question

Next Question