Find the value of 4- integration of sraccos xcos-sin x 3 d x within limits 0 to -2

J =4/π∫_0^π/2 cos x/cos x + sin xdx J = 4/π∫_0^π/2 sin x/sin x + cos xdx [sin θ = sin (a θ)] 2I = 4/π∫_0^π/2 sin x + cos x/sin x + cos xdx 2I = 4̸^2/π̸× ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

Find the valu...

Question

Find the value of $\frac{4}{π}$ (integration of \frac{cos x}{cos} + sin x 3dx within limits 0 to $\frac{π}{2}$ )

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Solution

$J = \frac{4}{π} \int_{0}^{\frac{π}{2}} \frac{c o s x}{c o s x + s i n x} d x J = \frac{4}{π} \int_{0}^{\frac{π}{2}} \frac{s i n x}{s i n x + c o s x} d x [s i n θ = s i n (a - θ)] 2 I = \frac{4}{π} \int_{0}^{\frac{π}{2}} \frac{s i n x + c o s x}{s i n x + c o s x} d x 2 I = \frac{/ 4^{2}}{/ π} \times \frac{/ π}{/ 2} \Rightarrow I = 1$

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Find the value of 4π (integration of \frac{cos x}{cos} + sin x 3dx within limits 0 to π2)

J=4π∫π20 cos xcos x+sin xdxJ=4π∫π20 sin xsin x+cos xdx [sin θ=sin (a−θ)]2I=4π∫π20 sin x+cos xsin x+cos xdx2I=/42/π×/π/2 ⇒I=1

Find the value of $\frac{4}{π}$ (integration of \frac{cos x}{cos} + sin x 3dx within limits 0 to $\frac{π}{2}$ )