Please give me the complete solution of 

Asked by Balbir | 23rd Oct, 2017, 08:50: PM

Expert Answer:

begin mathsize 16px style integral fraction numerator dx over denominator square root of sin cubed xcosx end root end fraction
Dividing space numerator space and space denominator space by space cos squared straight x
integral fraction numerator begin display style fraction numerator 1 over denominator space cos squared straight x end fraction end style dx over denominator begin display style fraction numerator begin display style 1 end style over denominator begin display style space cos squared straight x end style end fraction end style begin display style cross times end style square root of sin cubed xcosx end root end fraction
integral fraction numerator sec squared xdx over denominator square root of begin display style fraction numerator sin cubed xcosx over denominator cos to the power of 4 straight x end fraction end style end root end fraction
integral fraction numerator sec squared xdx over denominator square root of begin display style fraction numerator sin cubed straight x over denominator begin display style cos cubed end style begin display style straight x end style end fraction end style end root end fraction
integral fraction numerator sec squared xdx over denominator square root of begin display style tan cubed end style begin display style straight x end style end root end fraction
Put space tanx space equals space straight t space rightwards double arrow space sec squared xdx space equals space dt
integral dt over straight t to the power of begin display style 3 over 2 end style end exponent
Further space you space can space integrate. end style

Answered by Sneha shidid | 15th Dec, 2017, 11:05: AM