Correct Answer is : 142log⁡∣1+2sin⁡x1−2sin⁡x∣−14log⁡∣sec⁡x+tan⁡x∣+c{1}{4 {2}} ≤ft|{1+{2} x}{1-{2} x}|-1/4 | x+ x|+c421log1−2sinx1+2sinx−41log∣secx+tanx∣+c

{1}{4 {2}} ≤ft|{1+{2} x}{1-{2} x}|+1/4 | x+ x|+c

{1}{4 {2}} ≤ft|{1+{2} x}{1-{2} x}|-1/4 | x+ x|+c

{1}{4 {2}} ≤ft|{1-{2} x}{1+{2} x}|+1/4 | x+ x|+c

{1}{4 {2}} ≤ft|{1-{2} x}{1+{2} x}|-1/4 | x+ x|+c

Correct Answer is : 142log⁡∣1+2sin⁡x1−2sin⁡x∣−14log⁡∣sec⁡x+tan⁡x∣+c{1}{4 {2}} ≤ft|{1+{2} x}{1-{2} x}|-1/4 | x+ x|+c421log1−2sinx1+2sinx−41log∣secx+tanx∣+c

Test Index

MHTCET 11 Apr 2026 Shift 2 Paper

Question : 127 of 150

Marks: +1, -0

\int \frac{\sin x}{\sin 4 x} d x=

$\frac{1}{4 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x}\right|+\frac{1}{4} \log |\sec x+\tan x|+c$
$\frac{1}{4 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \sin x}{1-\sqrt{2} \sin x}\right|-\frac{1}{4} \log |\sec x+\tan x|+c$
$\frac{1}{4 \sqrt{2}} \log \left|\frac{1-\sqrt{2} \sin x}{1+\sqrt{2} \sin x}\right|+\frac{1}{4} \log |\sec x+\tan x|+c$
$\frac{1}{4 \sqrt{2}} \log \left|\frac{1-\sqrt{2} \sin x}{1+\sqrt{2} \sin x}\right|-\frac{1}{4} \log |\sec x+\tan x|+c$

Validate

Solution:

Concept:

Express

\sin 4x

using standard identities, simplify, substitute

t = \sin x

, and use partial fractions to integrate.

Explanation:

We know that:

\sin 4x = 4 \sin x \cos x \cos 2x

Therefore,

\frac{\sin x}{\sin 4x} = \frac{1}{4 \cos x \cos 2x}

The given integral becomes:

I = \frac{1}{4} \int \frac{dx}{\cos x \cos 2x}

Using

\cos 2x = 1 - 2\sin^2 x

, we get:

I = \frac{1}{4} \int \frac{dx}{\cos x(1 - 2\sin^2 x)}

Substitute

t = \sin x

, so

dt = \cos x \, dx

. Then:

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

Thus,

I = \frac{1}{4} \int \frac{dt}{(1 - t^2)(1 - 2t^2)}

Using partial fractions:

\frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{-1}{1 - t^2} + \frac{2}{1 - 2t^2}

So,

I = \frac{1}{4}\left[-\int \frac{dt}{1 - t^2} + 2\int \frac{dt}{1 - 2t^2}\right]

Now apply standard integrals:

\int \frac{dt}{1 - t^2} = \frac{1}{2} \log\left|\frac{1+t}{1-t}\right|

\int \frac{dt}{1 - 2t^2} = \frac{1}{2\sqrt{2}} \log\left|\frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right|

Substituting back:

I = -\frac{1}{8}\log\left|\frac{1+t}{1-t}\right| + \frac{1}{4\sqrt{2}}\log\left|\frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right| + c

Since

t = \sin x

, we have:

-\frac{1}{8}\log\left|\frac{1+\sin x}{1-\sin x}\right| = -\frac{1}{4}\log|\sec x + \tan x|

Therefore,

I = \frac{1}{4\sqrt{2}}\log\left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right| - \frac{1}{4}\log|\sec x + \tan x| + c

Answer:

Option B

\frac{1}{4\sqrt{2}} \log\left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right| - \frac{1}{4} \log|\sec x + \tan x| + c

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