Q91 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics

If $\int \dfrac{1}{\cos^3 x \sqrt{\sin 2x}}\, dx = p(\tan^2 x + q)\sqrt{\tan x} + c$, then the values of $p$ and $q$ respectively are

  • A
    $p = \dfrac{\sqrt{2}}{5}, q = \dfrac{1}{5}$
  • B
    $p = \dfrac{\sqrt{2}}{3}, q = 3$
  • Correct answer
    C
    $p = \dfrac{\sqrt{2}}{5}, q = 5$
  • D
    $p = \dfrac{2}{\sqrt{5}}, q = \sqrt{5}$

Solution

Let $I = \int \dfrac{1}{\cos^3 x \sqrt{\sin 2x}} dx$

Using $\sin 2x = 2 \sin x \cos x$, we get:

$I = \int \dfrac{1}{\cos^3 x \sqrt{2 \sin x \cos x}} dx$

$I = \dfrac{1}{\sqrt{2}} \int \dfrac{1}{\cos^{7/2} x \sin^{1/2} x} dx$

Dividing the numerator and the denominator by $\cos^4 x$, we get:

$I = \dfrac{1}{\sqrt{2}} \int \dfrac{\sec^4 x}{\tan^{1/2} x} dx$

Let $\tan x = t \Rightarrow \sec^2 x dx = dt$

$I = \dfrac{1}{\sqrt{2}} \int \dfrac{1 + t^2}{\sqrt{t}} dt$

$I = \dfrac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) dt$

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

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

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

$I = \dfrac{\sqrt{2}}{5} (t^2 + 5) \sqrt{t} + c$

Substituting $t = \tan x$:

$I = \dfrac{\sqrt{2}}{5} (\tan^2 x + 5) \sqrt{\tan x} + c$

Comparing this with $p(\tan^2 x + q)\sqrt{\tan x} + c$, we get:

$p = \dfrac{\sqrt{2}}{5}$ and $q = 5$

Answer: $p = \dfrac{\sqrt{2}}{5}, q = 5$

Hence, the correct answer is option: 'C'.

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