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Integrals

Question
CBSEENMA12032448

Evaluate  integral subscript 0 superscript straight pi over 6 end superscript space cos to the power of negative 3 end exponent 2 straight theta space sin space 2 straight theta space dθ

Solution

Let I = integral subscript 0 superscript straight pi over 6 end superscript space cos to the power of negative 3 end exponent 2 straight theta space sin 2 straight theta space space dθ space equals space integral subscript 0 superscript straight pi over 6 end superscript fraction numerator sin space 2 straight theta space dθ over denominator cos cubed space 2 straight theta end fraction
Put cos space 2 straight theta space equals space straight y comma space space therefore space space minus 2 space sin space 2 straight theta space dθ space equals space dy space space space rightwards double arrow space space sin space 2 straight theta space dθ space equals space minus 1 half dy
When straight theta space equals space 0 comma space space straight y space equals space cos space 0 space equals space 1
When straight theta space equals space straight pi over 6 comma space space space straight y space equals space cos straight pi over 3 space equals space 1 half
therefore        straight I space equals space minus 1 half integral subscript 1 superscript 1 half end superscript 1 over straight y cubed dy space equals space minus 1 half integral subscript 1 superscript 1 half end superscript straight y to the power of negative 3 end exponent space dy space equals space minus 1 half open square brackets fraction numerator straight y to the power of negative 2 end exponent over denominator negative 2 end fraction close square brackets to the power of 1 half end exponent
                equals space 1 fourth open square brackets 1 over straight y squared close square brackets subscript 1 superscript 1 half end superscript space equals space 1 fourth left square bracket 4 minus 1 right square bracket space equals space 1 fourth cross times 3 space equals space 3 over 4

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