Evaluate the following integral using substitution
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator 1 plus cos squared straight x end fraction dx$

Solution

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx space over denominator 1 plus cos squared straight x end fraction dx$
Put cos x = t, ∴ sin x dx = dt ⇒ sin x dx = – dt When x = 0, t = cos 0 = 1
When x = $straight pi over 2 comma space space straight t space equals space cos straight pi over 2 space equals space 0$
$therefore space space space space space space straight I space equals space minus integral subscript 1 superscript 0 fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space minus left square bracket tan to the power of negative 1 end exponent straight t right square bracket subscript 1 superscript 0 space equals space minus left square bracket tan to the power of negative 1 end exponent 0 minus tan to the power of negative 1 end exponent 1 right square bracket space equals space minus open square brackets 0 minus straight pi over 4 close square brackets space equals space straight pi over 4$

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Integrals

Evaluate the following integral using substitution
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator 1 plus cos squared straight x end fraction dx$

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