Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{4} \sin^3 2t\ \cos 2t\ d\ t . Then, \]
\[Let\ \sin 2t = u . Then, 2 \cos\ 2t\ dt = du\]
\[When\ t = 0, u = 0\ and\ t\ = \frac{\pi}{4}, u = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 u^3 du\]
\[ \Rightarrow I = \frac{1}{2} \left[ \frac{u^4}{4} \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{1}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{1}{8}\]
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"`
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
