Mathematics

Find :  $$\displaystyle\int \dfrac{sin x - cos x}{\sqrt{1 + sin \, 2x}} dx , \, 0 < x < \dfrac{\pi}{2}$$


SOLUTION
$$\sqrt{1+\sin 2{x}}=\sqrt{\sin^2 x+\cos ^2 x+2\sin x\cos x}\sqrt{(\sin x+\cos x)^2}=|\sin x+\cos x|$$
$$0<x<\dfrac{\pi}{2},\implies \sqrt{1+\sin 2{x}}=\sin x+\cos x$$
$$\displaystyle\int \dfrac{\sin x-\cos x}{\sqrt{1+\sin 2x}}d x=\int \dfrac{\sin x-\cos x}{\sin x+\cos x}d x=\int \dfrac{\tan x-1}{1+\tan x}d x=\int \tan \bigg(x-\dfrac{\pi}{4}\bigg)d x=\log \bigg|\text{sec}\bigg(x-\dfrac{\pi}{4}\bigg)\bigg|+C$$
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 Hard
Evaluate:
$$\int { \sqrt { { x }^{ 2 }-16 }  }  \ dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Hard
$$\int sec^{3}x dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Hard
Solve
$$\int { (3x-2)\sqrt { { x }^{ 2 }+x+1 } dx } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Solve $$\displaystyle\int {\dfrac{{{{\cos }^2}\theta d\theta }}{{{{\cos }^2}\theta  + 4{{\sin }^2}\theta }}} $$.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Medium
Let $$n \space\epsilon \space N$$ & the A.M., G.M., H.M. & the root mean square of $$n$$ numbers $$2n+1, 2n+2, ...,$$ up to $$n^{th}$$ number are $$A_{n}$$, $$G_{n}$$, $$H_{n}$$ and $$R_{n}$$ respectively. 
On the basis of above information answer the following questions

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer