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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{\pi / 4}^{3 \pi / 4} \frac{x}{1+\sin x} d x=$
  • ✓
    $\pi(\sqrt{2}-1)$
  • B
    $\pi(\sqrt{2}+1)$
  • C
    $2 \pi(\sqrt{2}-1)$
  • D
    $2 \pi(\sqrt{2}+1)$
✓

Answer

Correct option: A.
$\pi(\sqrt{2}-1)$
(A)
Let $I =\int_{\pi / 4}^{3 \pi / 4} \frac{x}{1+\sin x} d x$
$=\int_{\pi / 4}^{3 \pi / 4} \frac{x \sec x}{\sec x+\tan x} d x$
Let $I _1=\int \frac{\sec x}{\sec x+\tan x} d x$
Put $\frac{1}{\sec x+\tan x}= t$
$\Rightarrow-\frac{\left(\sec x \tan x+\sec ^2 x\right)}{(\sec x+\tan x)^2} d x= dt$
$\therefore \quad I _1=-\int \frac{-\sec x(\sec x+\tan x)}{(\sec x+\tan x)^2} d x$
= $-\int dt$
$\begin{array}{l}=- t + c \\ =\frac{-1}{\sec x+\tan x}+ c \end{array}$
$\therefore \quad I =\left[\frac{-x}{\sec x+\tan x}\right]_{\pi / 4}^{3 \pi / 4}-\int_{\pi / 4}^{3 \pi / 4} \frac{-1}{\sec x+\tan x} d x$
$\ldots\left[\because \int_{ a }^{ b }( uv ) d x=\left[ u \int vd x\right]_{ a }^{ b }-\int_{ a }^{ b }\left[\frac{ du }{ d x} \int v d x\right] d x\right]$
$=\left(\frac{\frac{-3 \pi}{4}}{-\sqrt{2}-1}\right)-\left(\frac{\frac{-\pi}{4}}{\sqrt{2}+1}\right)+\int_{\pi / 4}^{3 \pi / 4} \frac{\cos x}{1+\sin x} d x$
$=\frac{\pi}{1+\sqrt{2}}+[\log |1+\sin x|]_{\pi / 4}^{3 \pi / 4}$
$\begin{array}{l}=\frac{\pi}{1+\sqrt{2}}+\log \left|1+\frac{1}{\sqrt{2}}\right|-\log \left|1+\frac{1}{\sqrt{2}}\right| \\ =\frac{\pi}{1+\sqrt{2}}=\pi(\sqrt{2}-1)\end{array}$

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