If {1}{ ³ x { 2 x}} d x=p≤ft( ² x+q) { x}+c, then the values of p and q respectively are

Correct Answer is : p=25,q=5p={{2}}{5}, q=5p=52,q=5

Correct Answer is : p=25,q=5p={{2}}{5}, q=5p=52,q=5

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MHTCET 15 Apr 2026 Shift 1 Paper

Question : 126 of 150

Marks: +1, -0

\int \frac{1}{\cos ^{3} x \sqrt{\sin 2 x}} d x=p\left(\tan ^{2} x+q\right) \sqrt{\tan x}+c

p

q

respectively are

Solution:

Concept:

Substitute

t = \tan x

to transform the integral into a simple power form.

Explanation:

Let

t = \tan x

. Then

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

and

\sin 2x = \frac{2t}{1+t^2}

Also,

\sec^2 x = 1+t^2

, so

\frac{1}{\cos^3 x} = (1+t^2)^{3/2}

Rewrite the integral in terms of

t

I = \int \frac{(1+t^2)^{3/2}}{\sqrt{\frac{2t}{1+t^2}}} \cdot \frac{dt}{1+t^2}

Simplify:

I = \int \frac{1+t^2}{\sqrt{2t}}\,dt = \frac{1}{\sqrt{2}}\int \left(t^{-1/2} + t^{3/2}\right)dt

Integrate term by term:

I = \frac{1}{\sqrt{2}}\left(2t^{1/2} + \frac{2}{5}t^{5/2}\right) + c

I = \sqrt{2}\,t^{1/2} + \frac{\sqrt{2}}{5}t^{5/2} + c

Factor out

\frac{\sqrt{2}}{5}\sqrt{t}

I = \frac{\sqrt{2}}{5}\left(t^2 + 5\right)\sqrt{t} + c

Substitute back

t = \tan x

I = \frac{\sqrt{2}}{5}\left(\tan^2 x + 5\right)\sqrt{\tan x} + c

Compare with

p\left(\tan^2 x + q\right)\sqrt{\tan x} + c

Thus,

p = \frac{\sqrt{2}}{5}

and

q = 5

Answer:

p = \frac{\sqrt{2}}{5}

q = 5

, which is Option C.

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