MHB 7.2.1 integrate trig with radical

[karush]

View attachment 9277
Ok I tried to follow an example but
kinda got lost in this
maybe more steps

the answer is from W|A
View attachment 9278

 

[skeeter]

$$\int_0^{\pi/4} \sqrt{1+\cos(4x)} \, dx$$

$u=2x \implies du = 2 \, dx$

$$\frac{1}{2} \int_0^{\pi/4} \sqrt{1+\cos[2(2x)]} \cdot 2 \, dx$$

substitute and reset the limits of integration ...

$$\frac{1}{2} \int_0^{\pi/2} \sqrt{1+\cos(2u)} \, du$$

note the double angle identity, $\cos(2u) = 2\cos^2{u}-1 \implies 1+\cos(2u) = 2\cos^2{u}$

$$\frac{1}{2} \int_0^{\pi/2} \sqrt{2\cos^2{u}} \, du$$

note $\sqrt{2\cos^2{u}} = \sqrt{2} \cdot \sqrt{\cos^2{u}} = \sqrt{2} \cdot |\cos{u}| = \sqrt{2} \cdot \cos{u}$ since $\cos{u} \ge 0$ in quadrant I.

$$\frac{\sqrt{2}}{2} \int_0^{\pi/2}\cos{u} \, du$$

$\dfrac{\sqrt{2}}{2} \bigg[\sin{u} \bigg]_0^{\pi/2} = \dfrac{\sqrt{2}}{2} (1-0) = \dfrac{\sqrt{2}}{2}$

 

[karush]

ok I see now it was resetting the limits because of substitution

Mahalo