∫0π x1+sin⁡xdx is equal to

Let,I=∫0π x1+sin⁡xdx----iI=∫0π π−x1+sin⁡(π−x)dxthen, I=∫0π π−x1+sin⁡xdx ∵∫0a f(x)dx=∫0a f(a−x)dx [∵sin⁡(π−x)=sin⁡x]… (ii) On adding Eqs. (i) and (ii), we get2I=∫0π π(1+sin⁡x)dx=π∫0π 1(1+sin⁡x)dx=π∫0π 1−sin⁡x(1+sin⁡x)(1−sin⁡x)dx[multiply numerator and denominator by (1 - sin x)] ⇒ 2I=π∫0π 1−sin⁡x1−sin2⁡xdx =π∫0π 1cos2⁡xdx−π∫0π sin⁡xcos2⁡xdx⇒ 2I=π∫0π sec2⁡xdx−π∫0π sec⁡x⋅tan⁡xdx⇒ 2I=π[tan⁡x−sec⁡x]0π⇒ 2I=π[tan⁡π−sec⁡π−(tan⁡0−sec⁡0)]⇒ 2I=π[0+1−0+1]⇒2I=2π⇒I=π

∫0π x1+sin⁡xdx is equal to

Let,I=∫0π x1+sin⁡xdx----iI=∫0π π−x1+sin⁡(π−x)dxthen, I=∫0π π−x1+sin⁡xdx ∵∫0a f(x)dx=∫0a f(a−x)dx [∵sin⁡(π−x)=sin⁡x]… (ii) On adding Eqs. (i) and (ii), we get2I=∫0π π(1+sin⁡x)dx=π∫0π 1(1+sin⁡x)dx=π∫0π 1−sin⁡x(1+sin⁡x)(1−sin⁡x)dx[multiply numerator and denominator by (1 - sin x)] ⇒ 2I=π∫0π 1−sin⁡x1−sin2⁡xdx =π∫0π 1cos2⁡xdx−π∫0π sin⁡xcos2⁡xdx⇒ 2I=π∫0π sec2⁡xdx−π∫0π sec⁡x⋅tan⁡xdx⇒ 2I=π[tan⁡x−sec⁡x]0π⇒ 2I=π[tan⁡π−sec⁡π−(tan⁡0−sec⁡0)]⇒ 2I=π[0+1−0+1]⇒2I=2π⇒I=π

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

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$π$

$\frac{π}{2}$

$2 π$

$\frac{π}{8}$

answer is A.

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Detailed 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)]

$\begin{array}{l} \Rightarrow 2 I = π \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} dx \\ = π \int_{0}^{π} \frac{1}{\cos^{2} x} dx - π \int_{0}^{π} \frac{\sin x}{\cos^{2} x} dx \\ \Rightarrow 2 I = π \int_{0}^{π} \sec^{2} xdx - π \int_{0}^{π} \sec x \cdot \tan xdx \\ \Rightarrow 2 I = π [\tan x - \sec x]_{0}^{π} \\ \Rightarrow 2 I = π [\tan π - \sec π - (\tan 0 - \sec 0)] \\ \Rightarrow 2 I = π [0 + 1 - 0 + 1] \Rightarrow 2 I = 2 π \Rightarrow I = π \end{array}$