Question

Evaluate ∫₀^π √(1+cos8xdx)

Answer 1

Silution;

Substitute;

$u=8x$

So that ;

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

$dx=\frac 18du$

Substitute the values back to the equation;

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

Apply the half angle identity;

cos(u)+1=2cos²($\frac u2$ )

By substitution;

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

Simplify as;

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

Take;

$v=\frac u2$ ;$\frac{dv}{du}=\frac12$ ;$du=2dv$

By substitution;

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

Integrate;

$[\frac{\sqrt2}{4}sin(v)]_{0}^{4π}$

But $v=\frac u2$ ,also $u=8x$ ,hence $v=4x$

Replace back the value of v;

$[\frac{\sqrt2}{4}sin(4x)]_{0}^{4π}$

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