≤ft[ x (x+α)+ x (x+α)/ x (x+α) ] dx =

Correct Answer is : cot⁡αlog⁡(∣sin⁡xsin⁡(x+a)∣)+c α ≤ft( ≤ft| x/ (x+a) | )+ccotαlog(sin(x+a)sinx)+c

α ≤ft( ≤ft| x/ (x+a) | )+c

α ≤ft( ≤ft| x/ (x+α) | )+c

Correct Answer is : cot⁡αlog⁡(∣sin⁡xsin⁡(x+a)∣)+c α ≤ft( ≤ft| x/ (x+a) | )+ccotαlog(sin(x+a)sinx)+c

Test Index

TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

\int \left[ \frac{\cot x \cos (x+\alpha)+\sin x \sin (x+\alpha)}{\sin x \sin (x+\alpha)} \right] dx =

$\cot \alpha \log \left( \left| \frac{\sin x}{\sin (x+a)} \right| \right)+c$
$\log |\sin x \sin (x+a)|+x+c$
$\log |\sin x \cos (x+a)|+x+c$
$\tan \alpha \log \left( \left| \frac{\cos x}{\sin (x+\alpha)} \right| \right)+c$

Solution:

Consider the expression,

\int \left[ \frac{\cot x \cos (x+\alpha)+\sin x \sin (x+\alpha)}{\sin x \sin (x+\alpha)} \right] dx

= \int \left[ \frac{\cos \alpha}{\sin x \sin (x+\alpha)} \right] dx

= \frac{\cos \alpha}{\sin \alpha} \int \left[ \frac{\sin \alpha}{\sin x \sin (x+\alpha)} \right] dx

= \cos \alpha \int \left[ \frac{\sin (x+\alpha)-x}{\sin x \sin (x+\alpha)} \right] dx

Further simplify the above,

\int \left[ \frac{\cot x \cos (x+\alpha)+\sin x \sin (x+\alpha)}{\sin x \sin (x+\alpha)} \right] dx

= \cot \alpha \int \left[ \frac{\sin (x+\alpha) \cos x-\cos (x+\alpha) \sin x}{\sin x \sin (x+\alpha)} \right] dx

= \cot \alpha \int [\cot x-\cot (x+\alpha)] dx

= \cot \alpha [\log |\sin x|-\log |

\sin (x+\alpha)|]+ c

Further simplify the above,

\int \left[ \frac{\cot x \cos (x+\alpha)+\sin x \sin (x+\alpha)}{\sin x \sin (x+\alpha)} \right] dx

= \cot \alpha \log \left( \left| \frac{\sin x}{\sin (x+a)} \right| \right)+c

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