Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\cos x}{1 + \sin^2 x} d x . \]
\[Let\ \sin x\ = t\ . Then, \cos x\ dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{\cos x}{1 + \sin^2 x} d x\]
\[ \Rightarrow I = \int_0^1 \frac{1}{1 + t^2} d t\]
\[ \Rightarrow I = \left[ \tan^{- 1} t \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
Evaluate each of the following integral:
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
