Evaluate: { x}{1-4²x} dx

Correct Answer is : −14log⁡∣1+2cos⁡x1−2cos⁡x∣+C-1/4 ≤ft| 1+2 x/1-2 x | + C−41log1−2cosx1+2cosx+C

Correct Answer is : −14log⁡∣1+2cos⁡x1−2cos⁡x∣+C-1/4 ≤ft| 1+2 x/1-2 x | + C−41log1−2cosx1+2cosx+C

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UPSC NDA Math Model Paper 6

Question : 63 of 120

Marks: +1, -0

Evaluate:

\int \frac{\sin x}{1-4\cos^{2}x} dx

Solution:

Here we have to find the value of

\int \frac{\sin x}{1-4\cos^{2}x} dx

Let

\cos x = t

then

-\sin x dx = dt

\Rightarrow \int \frac{\sin x}{1-4\cos^{2}x} dx = -\int \frac{dt}{1-4t^{2}}

The integrand

\int \frac{dt}{1-4t^{2}}

can be written as:

\Rightarrow \int \frac{dt}{1-4t^{2}} = \frac{1}{4} \int \frac{dt}{\frac{1}{4} - t^{2}}

= \frac{1}{4}. \int \frac{dt}{\left(\frac{1}{2}\right)^{2} - t^{2}}

So, by comparing

\int \frac{dt}{\left(\frac{1}{2}\right)^{2} - t^{2}}

with

\int \frac{dx}{a^{2} - x^{2}}

we get:

a = \frac{1}{2}

As we know that,

\int \frac{dx}{a^{2} - x^{2}} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C

where

C

is a constant

\Rightarrow \int \frac{dt}{\left(\frac{1}{2}\right)^{2} - t^{2}} = \frac{1}{2 \times \frac{1}{2}}

. \log \left| \frac{\frac{1}{2}+t}{\frac{1}{2}-t} \right|

\Rightarrow -\int \frac{dx}{1-4t^{2}} = -\frac{1}{4} \log \left| \frac{1+2t}{1-2t} \right| + C

By substituting cos x = t in the above equation we get

\Rightarrow \int \frac{\sin x}{1-4\cos^{2}x} dx

= -\frac{1}{4} \log \left| \frac{1+2\cos x}{1-2\cos x} \right| + C

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