Q66 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics

$\int_\limits0^{\dfrac{\pi}{2}} \dfrac{\cos x}{1+\cos x+\sin x} d x=$

  • A
    $\dfrac{\pi}{4}+\dfrac{1}{2} \log 2$
  • B
    $\dfrac{\pi}{4}-\log 2$
  • C
    $\dfrac{\pi}{4}+\log 2$
  • Correct answer
    D
    $\dfrac{\pi}{4}-\dfrac{1}{2} \log 2$

Solution

Let $I = \displaystyle\int_0^{\pi/2} \dfrac{\cos x}{1 + \cos x + \sin x}dx$ ... (1)

Using $\displaystyle\int_0^a f(x)dx = \int_0^a f(a-x)dx$, substitute $x \to \dfrac{\pi}{2} - x$:

$I = \displaystyle\int_0^{\pi/2} \dfrac{\sin x}{1 + \sin x + \cos x}dx$ ... (2)

Adding (1) and (2):

$2I = \displaystyle\int_0^{\pi/2} \dfrac{\cos x + \sin x}{1 + \cos x + \sin x}dx = \int_0^{\pi/2} \left(1 - \dfrac{1}{1 + \cos x + \sin x}\right)dx$

$2I = \dfrac{\pi}{2} - J$ where $J = \displaystyle\int_0^{\pi/2} \dfrac{1}{1 + \cos x + \sin x}dx$

Using half-angle: $\cos x = 2\cos^2\dfrac{x}{2} - 1$, $\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$:

$J = \displaystyle\int_0^{\pi/2} \dfrac{1}{2\cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}dx$

Dividing by $2\cos^2\dfrac{x}{2}$:

$J = \displaystyle\int_0^{\pi/2} \dfrac{\frac{1}{2}\sec^2\frac{x}{2}}{1 + \tan\frac{x}{2}}dx$

Let $t = 1 + \tan\dfrac{x}{2}$, $dt = \dfrac{1}{2}\sec^2\dfrac{x}{2}dx$. Limits: $t: 1 \to 2$

$J = \displaystyle\int_1^2 \dfrac{1}{t}dt = \log 2$

$2I = \dfrac{\pi}{2} - \log 2 \Rightarrow I = \dfrac{\pi}{4} - \dfrac{1}{2}\log 2$

Hence, the correct answer is option: 'D'.

Relevant Questions