If \int dx \over \cos ^3x\sqrt 2\sin 2x = ( \tan x )^A + C( \tan x )^B + k, where k is a constant of integration, then A + B +C equals :

Indefinite Integrals PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q43 with Solution

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Correct answer: B

Given integral

We need to evaluate $I=\int \frac{dx}{\cos^3 x\,\sqrt{2\sin 2x}}.$

Simplify the expression inside the root

Using $\sin 2x=2\sin x\cos x,$ we get $2\sin 2x=4\sin x\cos x.$ Hence, $\sqrt{2\sin 2x}=\sqrt{4\sin x\cos x}=2\sqrt{\sin x\cos x}.$

So, $I=\int \frac{dx}{2\cos^3 x\sqrt{\sin x\cos x}}.$

Convert in terms of $\tan x$

Let $t=\tan x.$ Then $dx=\frac{dt}{1+t^2}, \qquad \sin x\cos x=\tan x\cos^2 x=t\cos^2 x.$ Also, $\cos^2 x=\frac{1}{1+t^2} \implies \cos x=\frac{1}{\sqrt{1+t^2}}.$ Thus, $\sqrt{\sin x\cos x}=\sqrt{t\cos^2 x}=\sqrt{t}\,\cos x=\frac{\sqrt t}{\sqrt{1+t^2}}.$

Now the denominator becomes

=2\cos^3 x\cdot \frac{\sqrt t}{\sqrt{1+t^2}}.$$ Since $$\cos^3 x=\frac{1}{(1+t^2)^{3/2}},$$ we get $$2\cos^3 x\sqrt{\sin x\cos x} =2\cdot \frac{1}{(1+t^2)^{3/2}}\cdot \frac{\sqrt t}{\sqrt{1+t^2}} =\frac{2\sqrt t}{(1+t^2)^2}.$$ Therefore, $$I=\int \frac{dx}{2\cos^3 x\sqrt{\sin x\cos x}} =\int \frac{\frac{dt}{1+t^2}}{\frac{2\sqrt t}{(1+t^2)^2}} =\int \frac{1+t^2}{2\sqrt t}\,dt.$$ 4. **Integrate** $$I=\frac12\int \left(t^{-1/2}+t^{3/2}\right)dt.$$ Now, $$\int t^{-1/2}dt=2t^{1/2},$$ $$\int t^{3/2}dt=\frac{2}{5}t^{5/2}.$$ So, $$I=\frac12\left(2t^{1/2}+\frac{2}{5}t^{5/2}\right)+k =t^{1/2}+\frac{1}{5}t^{5/2}+k.$$ Replacing $t=\tan x$, $$I=(\tan x)^{1/2}+\frac{1}{5}(\tan x)^{5/2}+k.$$ 5. **Match with the given form** The question states $$\int \frac{dx}{\cos^3 x\sqrt{2\sin 2x}}=(\tan x)^A + C(\tan x)^B + k.$$ Comparing, $$A=\frac12, \qquad B=\frac52, \qquad C=\frac15.$$ Therefore, $$A+B+C=\frac12+\frac52+\frac15 =3+\frac15 =\frac{16}{5}.$$ 6. **Option check** $$\frac{16}{5}$$ corresponds to **Option B**.

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View written solutionFree

Correct answer: B

Given integral

We need to evaluate $I=\int \frac{dx}{\cos^3 x\,\sqrt{2\sin 2x}}.$

Simplify the expression inside the root

Using $\sin 2x=2\sin x\cos x,$ we get $2\sin 2x=4\sin x\cos x.$ Hence, $\sqrt{2\sin 2x}=\sqrt{4\sin x\cos x}=2\sqrt{\sin x\cos x}.$

So, $I=\int \frac{dx}{2\cos^3 x\sqrt{\sin x\cos x}}.$

Convert in terms of $\tan x$

Now the denominator becomes

=2\cos^3 x\cdot \frac{\sqrt t}{\sqrt{1+t^2}}.$$ Since $$\cos^3 x=\frac{1}{(1+t^2)^{3/2}},$$ we get $$2\cos^3 x\sqrt{\sin x\cos x} =2\cdot \frac{1}{(1+t^2)^{3/2}}\cdot \frac{\sqrt t}{\sqrt{1+t^2}} =\frac{2\sqrt t}{(1+t^2)^2}.$$ Therefore, $$I=\int \frac{dx}{2\cos^3 x\sqrt{\sin x\cos x}} =\int \frac{\frac{dt}{1+t^2}}{\frac{2\sqrt t}{(1+t^2)^2}} =\int \frac{1+t^2}{2\sqrt t}\,dt.$$ 4. **Integrate** $$I=\frac12\int \left(t^{-1/2}+t^{3/2}\right)dt.$$ Now, $$\int t^{-1/2}dt=2t^{1/2},$$ $$\int t^{3/2}dt=\frac{2}{5}t^{5/2}.$$ So, $$I=\frac12\left(2t^{1/2}+\frac{2}{5}t^{5/2}\right)+k =t^{1/2}+\frac{1}{5}t^{5/2}+k.$$ Replacing $t=\tan x$, $$I=(\tan x)^{1/2}+\frac{1}{5}(\tan x)^{5/2}+k.$$ 5. **Match with the given form** The question states $$\int \frac{dx}{\cos^3 x\sqrt{2\sin 2x}}=(\tan x)^A + C(\tan x)^B + k.$$ Comparing, $$A=\frac12, \qquad B=\frac52, \qquad C=\frac15.$$ Therefore, $$A+B+C=\frac12+\frac52+\frac15 =3+\frac15 =\frac{16}{5}.$$ 6. **Option check** $$\frac{16}{5}$$ corresponds to **Option B**.

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