Advertisements
Advertisements
Question
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Advertisements
Solution
Using the property
`int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x`
Let I = `int_0^(pi/2) (sin^7x)/(sin^7x + cos^7x) "d"x` ........(1)
I = `int_0^(pi/2) (sin^7(pi/2 - x))/(sin^7(pi/2 - x) + cos^7(pi/2 - x)) "d"x`
I = `int_0^(pi/2) (cos^7x)/(cos^7x + sin^x) "d"x` .........(2)
Adding (1) and (2)
I + I = `int_0^(pi/2) (sin^7x + cos^7x)/(sin^7x + cos^7x "d"x`
2I `int_0^(pi/2) "d"x`
2I = `[x]_0^(pi/2) = [pi/2 - 0]`
2I = `pi/2`
⇒ I = `pi/4`
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
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
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Choose the correct alternative:
`Γ(3/2)`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
