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 integral:
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
`int_0^(2a)f(x)dx`
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
