Mathematics

Evaluate:
$$\displaystyle\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x}}{{\sin x + \cos x}}dx} $$


SOLUTION

Consider the given integral.

$$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\sin x}{\sin x+\cos x} \right)}dx$$    …… (1)

 

We know that

$$\int_{a}^{b}{f\left( x \right)}dx=\int_{a}^{b}{f\left( a+b-x \right)}dx$$

 

Therefore,

$$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\sin \left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)+\cos \left( \dfrac{\pi }{2}-x \right)} \right)}dx$$

$$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos x}{\cos x+\sin x} \right)}dx$$      ….. (2)

 

From equation (1) and (2), we get

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos x+\sin x}{\cos x+\sin x} \right)}dx $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{1}dx $$

$$ 2I=\left( x \right)_{0}^{\dfrac{\pi }{2}} $$

$$ 2I=\left( \dfrac{\pi }{2}-0 \right) $$

$$ I=\dfrac{\pi }{4} $$

 

Hence, the value is $$\dfrac{\pi }{4}$$.

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Evaluate the following integrals:$$\displaystyle \int {\dfrac{2}{\sqrt{x}-\sqrt{x+3}}.dx}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Hard
Evaluate the integral
$$\displaystyle \int_{0}^{\pi}\frac{ {x} {d} {x}}{4 {s}i {n}^{2} {x}+5}$$
  • A. $$\displaystyle \frac{\pi^{2}}{20}$$
  • B. $$\displaystyle \frac{\pi^{2}}{9}$$
  • C. $$\displaystyle \frac{\pi^{2}}{3\sqrt{5}}$$
  • D. $$\displaystyle \frac{\pi^{2}}{6\sqrt{5}}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Evaluate:
$$\displaystyle \int \dfrac{\sqrt{1 + x^2}}{x^4}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate : $$\displaystyle \int{\dfrac{1}{(x+1)^{2}}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Hard
Let $$\displaystyle 2I_{1}+I_{2}=\int \frac {e^{x}}{e^{2x}+e^{-2x}}dx$$  and  $$\displaystyle I_{1}+2I_{2}=\int \frac {e^{-x}}{e^{2x}+e^{-2x}}dx$$
On the basis of above information, answer the following questions :

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer