Advertisements
Advertisements
Question
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
Prove that:
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
