Question

By using the properties of definite integrals, evaluate the integral $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$.

Answer 1

Hint: To solve the question, we have to apply the appropriate definite integral properties to simplify the given integral. To solve further, apply trigonometric properties and integration properties to calculate the value of the given integral. We have to add the simplified terms of the integral to ease the procedure of solving.

Complete step-by-step answer:
Let the given integral be I
$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$ ….. (1)
We know the property of definite integrals which states that if $f(x)=\int_{0}^{a}{g(x)}dx$ then $f(x)=\int_{0}^{a}{g(a-x)}$
By applying the above property to the given integral, we get
$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}}dx$
By the trigonometric functions properties, we know that $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x,\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$
By applying the above property to the given integral, we get
$I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx$ ….. (2)
By adding equation (1) and equation (2), we get
$I+I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$

$2I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$
We know the property of definite integrals which states $\int_{0}^{a}{l(x)dx}+\int_{0}^{a}{m(x)}dx=\int_{0}^{a}{\left( l(x)+m(x) \right)dx}$
By applying the above property to the given integral, we get
$2I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx$
$2I=\int_{0}^{\dfrac{\pi }{2}}{\left( 1 \right)}dx$
We know the property of definite integrals which states $\int_{b}^{a}{cdx}=c(a-b)$where c is a constant.
By applying the above property to the given integral, we get

& 2I=1\left( \dfrac{\pi }{2}-0 \right) \\\ & 2I=\dfrac{\pi }{2} \\\ & I=\dfrac{\pi }{2}\times \dfrac{1}{2} \\\ & \Rightarrow I=\dfrac{\pi }{4} \\\ \end{aligned}$$ Thus, we get the value of $$\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$$ is equal to $$\dfrac{\pi }{4}$$ Note: The possibility of mistake can be, not the appropriate definite integral properties and trigonometric properties to simplify the given integral. The other possibility of mistake can be not adding the simplified terms instead of solving the expression directly, which will ease the procedure of solving. The alternative way of solving the given integral is by converting the trigonometric expression to algebraic expression and rationalising the given expression to calculate the answer.