COMEDK 2025 Morning Solved Paper

Question : 139 of 180

Marks: +1, -0

\int \tan ^2\left(5-\frac{x}{2}\right) d x=

Solution:

Concept:

Use the identity

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

and substitution to simplify the integral.

Explanation:

Let

u = 5 - \frac{x}{2}

Then

du = -\frac{1}{2}\,dx

, so

dx = -2\,du

The integral becomes

\int \tan^2\!\bigl(5-\tfrac{x}{2}\bigr)\,dx = \int \tan^2 u \cdot (-2\,du) = -2\int \tan^2 u\,du.

Replace

\tan^2 u

with

\sec^2 u - 1

-2\int (\sec^2 u - 1)\,du = -2\Bigl(\int \sec^2 u\,du - \int 1\,du\Bigr).

We know

\int \sec^2 u\,du = \tan u

, and

\int 1\,du = u

Thus,

-2(\tan u - u) + C = -2\tan u + 2u + C

Substitute back

u = 5 - \frac{x}{2}

-2\tan\!\Bigl(5-\tfrac{x}{2}\Bigr) + 2\Bigl(5-\tfrac{x}{2}\Bigr) + C = -2\tan\!\Bigl(5-\tfrac{x}{2}\Bigr) + 10 - x + C.

Combine the constants

10

and

C

into a new constant

C'

-2\tan\!\Bigl(5-\tfrac{x}{2}\Bigr) - x + C'.

Answer:

\displaystyle -2\tan\!\left(5-\frac{x}{2}\right) - x + c

(Option D)

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