Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\pi \cos^5 x d x\]
\[ = \int_0^\pi \cos x \left( \cos^2 x \right)^2 dx\]
\[ = \int_0^\pi \cos x \left( 1 - \sin^2 x \right)^2 dx\]
\[ Let \sin x = t, then\ \cos x\ dx = dt\]
\[When\, x \to 0 ; t \to 0\ and\ x \to \pi ; t \to 0\]
\[ \text{Therefore}, \]
\[I = \int_0^0 \left( 1 - t^2 \right)^2 dt\]
\[ = 0\]
APPEARS IN
RELATED QUESTIONS
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
Γ(1) is
