Advertisements
Advertisements
Question
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
Advertisements
Solution
\[\int_0^\frac{\pi}{2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos^2 x}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)^2} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x}{1 + \cos x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x - 1 + 1}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2 - \left( 1 + \cos x \right)}{\left( 1 + \cos x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{2}{1 + \cos x}dx - \int_0^\frac{\pi}{2} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{2\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx - \int_0^\frac{\pi}{2} dx\]
\[ = 2 \int_0^\frac{\pi}{2} \frac{1 - \cos x}{\sin^2 x}dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \int_0^\frac{\pi}{2} \left( \ cosec^2 x - \ cosec x\ cotx \right) dx - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 \left[ - cotx + \ cosec x \right]_0^\frac{\pi}{2} - \left[ x \right]_0^\frac{\pi}{2} \]
\[ = 2 - \frac{\pi}{2}\]
APPEARS IN
RELATED QUESTIONS
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
What does a definite integral represent?
What is an antiderivative of \[2x+1\]?
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
