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If \int \frac{\cos ^{2}x}{\sin ^{6}}dx=A\cot ^{5}x+B\cot ^{3}x+k, then A + B equals

  • Option 1)

     \frac{8}{15}

  • Option 2)

    \frac{-8}{15}

  • Option 3)

    \frac{15}{8}

  • Option 4)

    \frac{-15}{8}

 

Answers (1)

best_answer

As we learnt

Type of integration by substitution -

\int (f(x))^{n}\cdot f{}'(x)dx

\therefore \frac{\left [ f(x) \right ]^{n+1}}{n+1}+c

- wherein

Let f(x)=t

f{}'(x)dx=dt

 

 \int \frac{\cos^{2}x}{\sin ^{6}x}=\int \cot ^{2}x.cosec^{4}xdx

=\int \cot ^{2}x.(1+\cot^{2}x)cosec^{2}xdx

Put t=\cot x

\therefore I=\int t^{2}\left ( 1+t^{2} \right )\left ( -1 \right )dt=-\left [ \frac{t^{5}}{5}+\frac{t^{3}}{3} \right ]+k

\therefore A=-\frac{1}{5},B=-\frac{1}{3}

 


Option 1)

 \frac{8}{15}

Option 2)

\frac{-8}{15}

Option 3)

\frac{15}{8}

Option 4)

\frac{-15}{8}

Posted by

gaurav

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