The value of  \int_{0}^{\pi /2}\frac{sin^{3}x}{sinx+cos x}dx    is :

  • Option 1)

     \frac{\pi -2}{8}       

  • Option 2)

    \frac{\pi -1}{4}

  • Option 3)

     \frac{\pi -2}{4}

  • Option 4)

    \frac{\pi -1}{2}

 

Answers (1)

     I=\int_{0}^{\pi /2}\frac{sin^{3}\left ( x \right )}{sin\left ( x \right )+cos\left ( x \right )}dx

   I=\int_{0}^{\pi /2}\frac{sin^{3}\left ( \frac{\pi }{2}-x \right )}{sin\left ( \frac{\pi }{2}-x \right )+cos\left ( \frac{\pi }{2}-x \right )}dx=\int_{0}^{\pi /2}\frac{cos^{3}\left ( x \right )}{cos\left ( x \right )+sin\left ( x \right )}dx

2I=\int_{0}^{\pi /2}\frac{sin^{3}\left ( x \right )+cos^{3}\left ( x \right )}{sin\left ( x \right )+cos\left ( x \right )}dx

       =\int_{0}^{\pi /2}\left ( sin^{2}\left ( x \right )+cos^{2}\left ( x \right )-sin\left ( x \right )cos\left ( x \right ) \right )dx

        =\int_{0}^{\pi /2}\left ( \perp -\frac{1}{2}sin\left ( 2x \right ) \right )dx=\left [ x+\frac{cos\left ( 2x \right )}{4} \right ]^{\pi /2}_{0}

       =\left ( \frac{\pi }{2}-\frac{1}{4} \right )-\left ( 0+\frac{1}{4} \right )=\frac{\pi }{2}-\frac{1}{2}

   I=\frac{\pi -1}{4}

 

 

 


Option 1)

 \frac{\pi -2}{8}       

Option 2)

\frac{\pi -1}{4}

Option 3)

 \frac{\pi -2}{4}

Option 4)

\frac{\pi -1}{2}

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