Correct answer: (A) 1√2logtanx/2+(π)/8+C. Let I=∫dx/(sinx+cosx) =(1√2∫dx1√2sinx+1√2cosx=1√2∫dxcosπ4sinx+sinπ4cosx=1√2∫dxsinx+π4=1√2∫dx2sinx2+π8cosx2+π8=1)/(2√2)∫(sec^2x2+π8dx)/(tanx2+π8) Let tanx/2+(π)/8=t⇒1/2sec^2x/2+(π)/8dx=dt So, I=1/(2√2)∫(sec^2x2+π8dx)/(tanx2+π8)=1√2∫dt/t =1√2logt+C=1√2logtanx/2+(π)/8+C

Q115 / 160 · TS EAMCET 2021 (04 Aug Shift 1) · Mathematics

$\int \frac{d x}{\sin x + \cos x} =$

Correct answer
A
$\frac{1}{\sqrt{2}} \log |\tan (\frac{x}{2} + \frac{π}{8})| + C$
B
$\frac{1}{\sqrt{2}} \log |\tan (\frac{x}{2} + \frac{π}{4})| + C$
C
$\frac{1}{\sqrt{2}} \log |\tan (\frac{x}{4} + \frac{π}{2})| + C$
D
$\frac{1}{\sqrt{2}} \log |\tan (\frac{x}{8} + \frac{π}{2})| + C$

Solution

Let $I = \int \frac{d x}{\sin x + \cos x}$

$= \frac{1}{\sqrt{2}} \int \frac{d x}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x}$

$= \frac{1}{\sqrt{2}} \int \frac{d x}{\cos \frac{π}{4} \sin x + \sin \frac{π}{4} \cos x}$

$= \frac{1}{\sqrt{2}} \int \frac{d x}{\sin (x + \frac{π}{4})}$

$= \frac{1}{\sqrt{2}} \int \frac{d x}{2 \sin (\frac{x}{2} + \frac{π}{8}) \cos (\frac{x}{2} + \frac{π}{8})}$

$= \frac{1}{2 \sqrt{2}} \int \frac{\sec^{2} (\frac{x}{2} + \frac{π}{8}) d x}{\tan (\frac{x}{2} + \frac{π}{8})}$

Let $\tan (\frac{x}{2} + \frac{π}{8}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2} + \frac{π}{8}) d x = d t$

So, $I = \frac{1}{2 \sqrt{2}} \int \frac{\sec^{2} (\frac{x}{2} + \frac{π}{8}) d x}{\tan (\frac{x}{2} + \frac{π}{8})} = \frac{1}{\sqrt{2}} \int \frac{d t}{t}$

$= \frac{1}{\sqrt{2}} \log |t| + C = \frac{1}{\sqrt{2}} \log |\tan (\frac{x}{2} + \frac{π}{8})| + C$

Hence, the correct answer is option: 'A'.

∫dxsinx+cosx=

Solution

Relevant Questions

$\int \frac{d x}{\sin x + \cos x} =$