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The value of   I=\int_{0}^{\pi /2}\frac{(sinx+cosx)^{2}}{\sqrt{1+sin2x}}\, dx\; \; \; is

  • Option 1)

    2

  • Option 2)

    1

  • Option 3)

    0

  • Option 4)

    3

 

Answers (1)

best_answer

As learnt in concept

Integrals for Trigonometric functions -

\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos x \right ) =sinx

\therefore \int sinxdx=-cosx+c

-

 

 

lower and upper limit -

\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}

                = F\left ( b \right )-F\left ( a \right )

 

- wherein

Where a is lower and b is upper limit.

 

 I=\int_{o}^{\frac{\pi }{2}}\frac{(sinx+cosx)^{2}}{\sqrt{1+sin2x}} dx

Now 1+sin2x=sin^{2}x+cos^{2}x+2sinxcosx

=(sinx+cosx)^{2}

We get I=\int_{o}^{\frac{\pi }{2}}\frac{(sinx+cosx)^{2}}{(sinx+cosx)}dx

I=\int_{o}^{\frac{\pi }{2}}(sinx+cosx)dx

=(-cosx)_{0}^{\frac{\pi }{2}}+(sinx)_{0}^{\frac{\pi }{2}}

=2


Option 1)

2

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Option 2)

1

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Option 3)

0

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Option 4)

3

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