Advertisements
Advertisements
Question
Advertisements
Solution
\[I = \int_0^\frac{\pi}{2} \cos^4 x \cos\ x\ d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right)^2 \cos x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - 2 \sin^2 x + \sin^4 x \right) \cos x dx \]
\[Let\ \sin x = t . Then, \cos dx = du\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[ \therefore I = \int_0^1 \left( 1 - 2 t^2 + t^4 \right) dt\]
\[ \Rightarrow I = \left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right]_0^1 \]
\[ \Rightarrow I = 1 - \frac{2}{3} + \frac{1}{5}\]
\[ \Rightarrow I = \frac{8}{15}\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate each of the following integral:
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
