{cosec}^{-1}≤ft({a+x/x}) d x=

Correct Answer is : aθtan⁡2θ−atan⁡θ+aθ+c(a θ ² θ-a θ+a θ+c≤ft(.aθtan2θ−atanθ+aθ+c( where θ=tan⁡−1(xa))≤ft. θ= ^{-1}≤ft({x/a}))θ=tan−1(ax)) and c is the constant of integration

a θ ² θ-a θ-a θ+c≤ft(. where ≤ft. θ= ^{-1}≤ft({x/a})) and c is the constant of integration

a θ ² θ-a θ+a θ+c≤ft(. where ≤ft. θ= ^{-1}≤ft({x/a})) and c is the constant of integration

a θ ² θ+a θ-a θ+c≤ft(. where ≤ft. θ= ^{-1}≤ft({x/a})) and c is the constant of integration

a θ ² θ+a θ+a θ+c≤ft(. where ≤ft. θ= ^{-1}≤ft({x/a})) and c is the constant of integration

{cosec}^{-1}≤ft({a+x/x}) d x=

Correct Answer is : aθtan⁡2θ−atan⁡θ+aθ+c(a θ ² θ-a θ+a θ+c≤ft(.aθtan2θ−atanθ+aθ+c( where θ=tan⁡−1(xa))≤ft. θ= ^{-1}≤ft({x/a}))θ=tan−1(ax)) and c is the constant of integration

Test Index

MHTCET 15 Apr 2026 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

\int \operatorname{cosec}^{-1}\left(\sqrt{\frac{a+x}{x}}\right) d x=

$a \theta \tan ^{2} \theta-a \tan \theta-a \theta+c\left(\right.$ where $\left.\quad \theta=\tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)\right)$ and c is the constant of integration
$a \theta \tan ^{2} \theta-a \tan \theta+a \theta+c\left(\right.$ where $\left.\quad \theta=\tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)\right)$ and c is the constant of integration
$a \theta \tan ^{2} \theta+a \tan \theta-a \theta+c\left(\right.$ where $\left.\quad \theta=\tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)\right)$ and c is the constant of integration
$a \theta \tan ^{2} \theta+a \tan \theta+a \theta+c\left(\right.$ where $\left.\quad \theta=\tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)\right)$ and c is the constant of integration

Validate

Solution:

Concept:

Use substitution to simplify the inverse cosecant expression, then apply integration by parts.

Explanation:

Let

\theta=\tan^{-1}\left(\sqrt{\frac{x}{a}}\right)

Then

\tan\theta=\sqrt{\frac{x}{a}}

, so squaring gives

x=a\tan^{2}\theta

Simplify the integrand:

\sqrt{\frac{a+x}{x}}=\sqrt{\frac{a+a\tan^{2}\theta}{a\tan^{2}\theta}}=\sqrt{\frac{1+\tan^{2}\theta}{\tan^{2}\theta}}

Using

1+\tan^{2}\theta=\sec^{2}\theta

, we get:

\sqrt{\frac{\sec^{2}\theta}{\tan^{2}\theta}}=\sqrt{\frac{1}{\sin^{2}\theta}}=\csc\theta

Hence

\operatorname{cosec}^{-1}\left(\sqrt{\frac{a+x}{x}}\right)=\theta

So the integral becomes:

\int \theta\,dx

Using integration by parts:

\int \theta\,dx=x\theta-\int x\,d\theta

Since

x=a\tan^{2}\theta

\int \theta\,dx=a\theta\tan^{2}\theta-a\int\tan^{2}\theta\,d\theta

Now

\tan^{2}\theta=\sec^{2}\theta-1

, therefore:

\int\tan^{2}\theta\,d\theta=\int(\sec^{2}\theta-1)\,d\theta=\tan\theta-\theta

Thus:

\int \theta\,dx=a\theta\tan^{2}\theta-a(\tan\theta-\theta)+C

=a\theta\tan^{2}\theta-a\tan\theta+a\theta+C

Answer:

Option B:

a\theta\tan^{2}\theta-a\tan\theta+a\theta+C

, where

\theta=\tan^{-1}\left(\sqrt{\frac{x}{a}}\right)

and

C

is the constant of integration.

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