Question

Evaluate :$\int \frac{[\cos x-\sin x]}{[1+2sinxcosx]}dx$

• A. $\left[\frac{-1}{(\cos x–\sin x)}\right]+c$
• B. $\frac{[\cos x+\sin x]}{[\cos x–\sin x]}+c$
• C. $\left[\frac{-1}{(\sin x+\cos x)}\right]+c$
• D. $\left[\frac{-x}{(\sin x+\cos x)}\right]+c$

Answer 1

The correct option is C

$\left[\frac{-1}{(\sin x+\cos x)}\right]+c$

Explanation for the correct option:

Finding $\int \frac{[\cos x-\sin x]}{[1+2sinxcosx]}dx$:

Let $I=\int \frac{[\cos x-\sin x]}{[1+2sinxcosx]}dx$

We know that

$\begin{array}{rcl}1+2\sin x\cos x& =& {\left(\sin x+\cos x\right)}^2\end{array}$

Substituting this identity we get

$\begin{array}{rcl}I& =& \int \frac{(\cos x-\sin x)}{{(\sin x+\cos x)}^2}dx\end{array}$

Substituting $\sin x+\cos x=t$

Differentiating this we get,

$(\cos x-\sin x)dx=dt$

Substituting the values of $t$ and $dt$, we get

$\begin{array}{rcl}\therefore I& =& \int \frac{1}{t^2}dt\\ I& =& -\frac{1}{t}+c;\left[\int x^{n}dx=\frac{x^{n+1}}{n+1}\right]\\ & =& -\frac{1}{\sin x+\cos x}+c;[t=\sin x+\cos x]\end{array}$

Hence, option (C) is the correct answer.