First slide

Evaluation of definite integrals

Question

$\int_{0}^{π} \frac{x}{1 + \sin x} dx$ is equal to

Moderate

A

$π$
B

$\frac{π}{2}$
C

$2 π$
D

$\frac{π}{8}$

Solution

Let,
$\begin{array}{l} I = \int_{0}^{π} \frac{x}{1 + \sin x} dx - - - - (i) \\ I = \int_{0}^{π} \frac{π - x}{1 + \sin (π - x)} dx \\ then, I = \int_{0}^{π} \frac{π - x}{1 + \sin x} dx [∵ \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx] \\ [∵ \sin (π - x) = \sin x] \dots (ii) \end{array}$
On adding Eqs. (i) and (ii), we get
$\begin{matrix} 2 I = \int_{0}^{π} \frac{π}{(1 + \sin x)} dx = π \int_{0}^{π} \frac{1}{(1 + \sin x)} dx \\ = π \int_{0}^{π} \frac{1 - \sin x}{(1 + \sin x) (1 - \sin x)} dx \end{matrix}$
[multiply numerator and denominator by (1 - sin x)]
$\begin{array}{l} \Rightarrow 2 I = π \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} dx \\ = π \int_{0}^{π} \frac{1}{\cos^{2} x} dx - π \int_{0}^{π} \frac{\sin x}{\cos^{2} x} dx \\ \Rightarrow 2 I = π \int_{0}^{π} \sec^{2} xdx - π \int_{0}^{π} \sec x \cdot \tan xdx \\ \Rightarrow 2 I = π [\tan x - \sec x]_{0}^{π} \\ \Rightarrow 2 I = π [\tan π - \sec π - (\tan 0 - \sec 0)] \\ \Rightarrow 2 I = π [0 + 1 - 0 + 1] \Rightarrow 2 I = 2 π \Rightarrow I = π \end{array}$

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