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Integrals

Question
CBSEENMA12035723

Evaluate:
integral fraction numerator 1 over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction dx

Solution

We need to evaluate integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction
Let space straight I space equals integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction
Multiply the numerator and the denominator by sec to the power of 4 straight x, we have
straight I space equals space integral fraction numerator sec to the power of 4 xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction
straight I space equals space integral fraction numerator sec squared straight x cross times sec squared xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction
Now space substitute space straight t space equals space tanx semicolon space dt space equals space sec squared xdx
Therefore comma space
straight I space equals space integral fraction numerator left parenthesis 1 plus straight t squared right parenthesis dt over denominator straight t to the power of 4 plus straight t squared plus 1 end fraction
straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style plus 1 close parentheses end fraction
straight I space equals integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style minus 2 plus 2 plus 1 close parentheses end fraction
straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t minus begin display style 1 over straight t end style close parentheses squared plus 3 end fraction
 Substitute space straight z equals straight t minus 1 over straight t semicolon space space dz space equals space open parentheses 1 plus 1 over straight t squared close parentheses dt
straight I space equals space integral fraction numerator dz over denominator straight z squared plus 3 end fraction
straight I space equals space integral fraction numerator dz over denominator straight z squared plus open parentheses square root of 3 close parentheses squared end fraction
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight z over denominator square root of 3 end fraction close parentheses plus straight c
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t minus begin display style 1 over straight t end style over denominator square root of 3 end fraction close parentheses plus straight c
straight I equals space fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus begin display style 1 over tanx end style over denominator square root of 3 end fraction close parentheses plus straight c
straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus cotx over denominator square root of 3 end fraction close parentheses plus straight c

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