SATHEE: Indefinite Integration 1 Question 5

Indefinite Integration 1 Question 5

6. The value of $\int_{0}^{π / 2} \frac{\sin^{3} x}{\sin x + \cos x} d x$ is

(a) $\frac{π - 1}{2}$

(b) $\frac{π - 2}{8}$

(d) $\frac{π - 2}{4}$

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Solution:

Key Idea Use property of definite integral.

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

Let

$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} x}{\sin x + \cos x} d x$

On applying the property, $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ , we get

$I = \int_{0}^{π / 2} \frac{\cos^{3} x}{\cos x + \sin x} d x$

On adding integrals (i) and (ii), we get

$\begin{aligned} 2 I & = \int_{0}^{π / 2} \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x} d x \\ = \int_{0}^{\frac{π}{2}} \frac{(\sin x + \cos x) (\sin^{2} x + \cos^{2} x - \sin x \cos x)}{\sin x + \cos x} d x \\ = \int_{0}^{\frac{π}{2}} 1 - \frac{1}{2} (2 \sin x \cos x) d x \end{aligned}$

$= \int_{0}^{\frac{π}{2}} 1 - \frac{1}{2} \sin 2 x d x$

$\begin{aligned} = x + \frac{1}{4} \cos 2 x_{0}^{π / 2} = \frac{π}{2} - 0 + \frac{1}{4} (- 1 - 1) = \frac{π}{2} - \frac{1}{2} \\ \Rightarrow I & = \frac{π}{4} - \frac{1}{4} = \frac{π - 1}{4} \end{aligned}$

Indefinite Integration 1 Question 5

6. The value of $\int_{0}^{π / 2} \frac{\sin^{3} x}{\sin x + \cos x} d x$ is

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Indefinite Integration 1 Question 5

6. The value of ∫0π/2sin3⁡xsin⁡x+cos⁡xdx is

Solution:

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6. The value of $\int_{0}^{π / 2} \frac{\sin^{3} x}{\sin x + \cos x} d x$ is