The value of d x/5+4 x is equal to

Correct Answer is : 23tan⁡−1(5tan⁡x2+43)+c2/3 ^{-1}≤ft({5 x/2+4}{3})+c32tan−1(35tan2x+4)+c, (where c is a constant of integration)

2/5 ^{-1}≤ft({5 x/2+4}{3})+c, (where c is a constant of integration)

2/3 ^{-1}≤ft({5 x/2+4}{3})+c, (where c is a constant of integration)

2/5 ≤ft({5 x/2+7}{5 x/2+1})+c, (where c is a constant of integration)

2/3 ≤ft({5 x/2+7}{5 x/2+1})+c, (where c is a constant of integration)

Correct Answer is : 23tan⁡−1(5tan⁡x2+43)+c2/3 ^{-1}≤ft({5 x/2+4}{3})+c32tan−1(35tan2x+4)+c, (where c is a constant of integration)

Test Index

MHTCET 15 May 2024 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

The value of

\int \frac{d x}{5+4 \sin x}

is equal to

$\frac{2}{5} \tan ^{-1}\left(\frac{5 \tan \frac{x}{2}+4}{3}\right)+c$ , (where c is a constant of integration)
$\frac{2}{3} \tan ^{-1}\left(\frac{5 \tan \frac{x}{2}+4}{3}\right)+c$ , (where c is a constant of integration)
$\frac{2}{5} \log \left(\frac{5 \tan \frac{x}{2}+7}{5 \tan \frac{x}{2}+1}\right)+c$ , (where c is a constant of integration)
$\frac{2}{3} \log \left(\frac{5 \tan \frac{x}{2}+7}{5 \tan \frac{x}{2}+1}\right)+c$ , (where c is a constant of integration)

Solution:

Concept:

Use the half-angle substitution

t = \tan\left(\frac{x}{2}\right)

to convert the integral into the standard form

\int \frac{du}{u^2+a^2}

Explanation:

Let

I = \int \frac{dx}{5+4\sin x}

Put

t = \tan\left(\frac{x}{2}\right)

Then

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

and

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

Substitute these into the integral:

I = \int \frac{\frac{2\,dt}{1+t^2}}{5+4\left(\frac{2t}{1+t^2}\right)}

Simplify the denominator:

I = 2\int \frac{dt}{5t^2+8t+5}

Factor out

5

and complete the square:

I = \frac{2}{5}\int \frac{dt}{t^2+\frac{8}{5}t+1}

I = \frac{2}{5}\int \frac{dt}{\left(t+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}

Use the formula

\int \frac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+c

I = \frac{2}{5}\cdot \frac{1}{\frac{3}{5}}\tan^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)+c

Simplify:

I = \frac{2}{3}\tan^{-1}\left(\frac{5t+4}{3}\right)+c

Replace

t

with

\tan\left(\frac{x}{2}\right)

I = \frac{2}{3}\tan^{-1}\left(\frac{5\tan\left(\frac{x}{2}\right)+4}{3}\right)+c

Answer:

Option B:

\frac{2}{3} \tan^{-1}\left(\frac{5 \tan \frac{x}{2}+4}{3}\right)+c

, where

c

is the constant of integration.

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