Answer to Question #223280 in Calculus for wgarnster

Question #223280

Evaluate ∫0π √(1+cos8xdx)


1
Expert's answer
2021-10-27T14:48:33-0400

Silution;

Substitute;

u=8xu=8x

So that ;

dudx=8\frac{du}{dx}=8

dx=18dudx=\frac 18du

Substitute the values back to the equation;

1808π1+cos(u)du\frac18\int_{0}^{8π}\sqrt{1+cos(u)}du

Apply the half angle identity;

cos(u)+1=2cos2(u2\frac u2 )

By substitution;

1808π2cos2(u2)\frac18\int_0^{8π}\sqrt2\sqrt{cos^2(\frac u2)}

Simplify as;

1808π2cos(u2)du\frac 18\int_{0}^{8π}\sqrt{2}cos(\frac u2)du

Take;

v=u2v=\frac u2 ;dvdu=12\frac{dv}{du}=\frac12 ;du=2dvdu=2dv

By substitution;

2404πcos(v)dv\frac{\sqrt{2}}{4}\int_{0}^{4π}cos(v)dv

Integrate;

[24sin(v)]04π[\frac{\sqrt2}{4}sin(v)]_{0}^{4π}

But v=u2v=\frac u2 ,also u=8xu=8x ,hence v=4xv=4x

Replace back the value of v;

[24sin(4x)]04π[\frac{\sqrt2}{4}sin(4x)]_{0}^{4π}

24[sin(4π)sin(0)]\frac{\sqrt2}{4}[sin(4π)-sin(0)] =0=0










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