Question

By using the properties of definite integrals,evaluate the integrals: $\int^\frac{\pi}{2}_0 \frac{cos^{5}\ x\ dx}{sin^{5}\ x+cos^{5}\ x} \, dx$

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

$\bf\int^\frac{\pi}{2}_0 \frac{cos^{5}x}{sin^{5}x+cos^{5}x}\:dx=\frac{\pi}{4}$

Step-by-step explanation:

Let

$I=\int^\frac{\pi}{2}_0 \frac{cos^{5}x}{sin^{5}x+cos^{5}x}\:dx$.........(1)

Using the properties of definite integrals

$\boxed{\int\limits^a_0\:f(x)\:dx=\int\limits^a_0\:f(a-x)\:dx}$

$I=\int^\frac{\pi}{2}_0 \frac{cos^{5}(\frac{\pi}{2}-x)}{sin^{5}(\frac{\pi}{2}-x)+cos^{5}(\frac{\pi}{2}-x)}\:dx$

$I=\int^\frac{\pi}{2}_0 \frac{sin^{5}x}{cos^{5}x+sin^{5}x}\:dx$

$I=int^\frac{\pi}{2}_0 \frac{sin^{5}x}{sin^{5}x+cos^{5}x}\:dx$..........(2)

Adding (1) and (2)

$2I=\int^\frac{\pi}{2}_0 \frac{cos^{5}x}{sin^{5}x+cos^{5}x}\:dx+\int^\frac{\pi}{2}_0 \frac{sin^{5}x}{sin^{5}x+cos^{5}x}\:dx$

$\implies\:2I=\int^\frac{\pi}{2}_0 \frac{cosx^{5}x+sin^{5}x}{sin^{5}x+cos^{5}x}\:dx$

$\implies\:2I=\int^\frac{\pi}{2}_0\:dx$

$\implies\:2I=[x]^\frac{\pi}{2}_0$

$\implies\:2I=\frac{\pi}{2}-0$

$\implies\:2I=\frac{\pi}{2}$

$\implies\:I=\frac{\pi}{4}$

$\implies\boxed{\bf\int^\frac{\pi}{2}_0 \frac{cos^{5}x}{sin^{5}x+cos^{5}x}\:dx=\frac{\pi}{4}}$