Q113 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
$\int \tan ^2\left(5-\dfrac{x}{2}\right) d x=$
- A$-2 \tan \left(5-\dfrac{x}{2}\right)+c$
- Correct answerB$-2 \tan \left(5-\dfrac{x}{2}\right)-x+c$
- C$\tan \left(5-\dfrac{x}{2}\right)+c$
- D$-\dfrac{1}{2} \tan \left(5-\dfrac{x}{2}\right)-x+c$
Solution
Let $I = \int \tan^{2}\left(5 - \dfrac{x}{2}\right) dx$.
Using the trigonometric identity $\tan^{2}\theta = \sec^{2}\theta - 1$, we have:
$I = \int \left[ \sec^{2}\left(5 - \dfrac{x}{2}\right) - 1 \right] dx$
Integrating term by term:
$I = \int \sec^{2}\left(5 - \dfrac{x}{2}\right) dx - \int 1 dx$
For the first integral, let $u = 5 - \dfrac{x}{2}$, then $du = -\dfrac{1}{2} dx$, which implies $dx = -2 du$.
$\int \sec^{2}\left(5 - \dfrac{x}{2}\right) dx = -2 \int \sec^{2} u du = -2 \tan u = -2 \tan\left(5 - \dfrac{x}{2}\right)$
Therefore, $I = -2 \tan\left(5 - \dfrac{x}{2}\right) - x + c$.
Answer: $-2 \tan \left(5-\dfrac{x}{2}\right)-x+c$
Using the trigonometric identity $\tan^{2}\theta = \sec^{2}\theta - 1$, we have:
$I = \int \left[ \sec^{2}\left(5 - \dfrac{x}{2}\right) - 1 \right] dx$
Integrating term by term:
$I = \int \sec^{2}\left(5 - \dfrac{x}{2}\right) dx - \int 1 dx$
For the first integral, let $u = 5 - \dfrac{x}{2}$, then $du = -\dfrac{1}{2} dx$, which implies $dx = -2 du$.
$\int \sec^{2}\left(5 - \dfrac{x}{2}\right) dx = -2 \int \sec^{2} u du = -2 \tan u = -2 \tan\left(5 - \dfrac{x}{2}\right)$
Therefore, $I = -2 \tan\left(5 - \dfrac{x}{2}\right) - x + c$.
Answer: $-2 \tan \left(5-\dfrac{x}{2}\right)-x+c$
Hence, the correct answer is option: 'B'.