Calcule
(a)
\int \operatorname{sen}^{3} x \sqrt{\cos x} dx
(b) \int_{0}^{\pi} \operatorname{sen}^{2} \frac{t}{2} \cos ^{5} \frac{t}{2} dt
(c) \int \operatorname{sen}^{2} t \cos ^{2} t dt
(d) \int \frac{\operatorname{sec} ^{4} t}{\tan t} dt
(e) \int_{0}^{\frac{\pi}{3}} \tan ^{2} x \operatorname{sec} x dx

Passo 1

O enunciado pede cinco integrais envolvendo potências de funções trigonométricas. Vamos usar identidades como \sin^2x=1-\cos^2x, \tan^2x=\operatorname{sec}^2x-1, a substituição u=\cos x ou u=\tan x, e, quando necessário, a redução de potências (por exemplo, escrever \sin^2t\cos^2t em termos de \cos4t).

Passo 2

Para \displaystyle\int \sin^3x\sqrt{\cos x}\,dx escrevemos \sin^3x=\sin x\,(1-\cos^2x) e fazemos u=\cos x, de modo que du=-\sin x\,dx. Assim
\int \sin^3x\sqrt{\cos x}\,dx =\,-\int(1-u^2)\,u^{\tfrac12}\,du =-\int\bigl(u^{\tfrac12}-u^{\tfrac52}\bigr)\,du.
Integramos em u:
-\Bigl[\tfrac23u^{\tfrac32}-\tfrac27u^{\tfrac72}\Bigr]+C =-\tfrac23\cos^{\tfrac32}x+\tfrac27\cos^{\tfrac72}x+C.

Passo 3

Para \displaystyle\int_{0}^{\pi}\sin^2\frac t2\cos^5\frac t2\,dt fazemos u=\tfrac t2, dt=2\,du, e os limites vão de 0 a \pi/2. A integral vira
2\int_{0}^{\frac\pi2}\sin^2u\,\cos^5u\,du =2\int_{0}^{\frac\pi2}(1-\sin^2u)\cos^5u\,du =2\Bigl(\int_{0}^{\frac\pi2}\cos^5u\,du-\int_{0}^{\frac\pi2}\cos^7u\,du\Bigr).
Usando a substituição w=\sin u, cada integral de \cos^n reduz a polinômio em w. Calcula-se \int_{0}^{\frac\pi2}\cos^5u\,du=8/15 e \int_{0}^{\frac\pi2}\cos^7u\,du=16/35.
Logo
2\Bigl(\tfrac{8}{15}-\tfrac{16}{35}\Bigr) =2\cdot\tfrac{8}{105} =\tfrac{16}{105}.

Passo 4

Para \displaystyle\int \sin^2t\cos^2t\,dt usamos a identidade de ângulo duplo:
\sin^2t\cos^2t=\tfrac14\sin^2(2t) =\tfrac14\cdot\tfrac{1-\cos4t}{2} =\tfrac18(1-\cos4t).
Então
\int \sin^2t\cos^2t\,dt =\tfrac18\int(1-\cos4t)\,dt =\tfrac18\Bigl(t-\tfrac{\sin4t}{4}\Bigr)+C.

Passo 5

Para \displaystyle\int \frac{\operatorname{sec}^4t}{\tan t}\,dt, fazemos u=\tan t, logo du=\operatorname{sec}^2t\,dt. Escrevemos
\frac{\operatorname{sec}^4t}{\tan t}\,dt =\frac{\operatorname{sec}^2t}{\tan t}\,\operatorname{sec}^2t\,dt =\frac{1+u^2}{u}\,du =\Bigl(\tfrac1u+u\Bigr)du,
pois \operatorname{sec}^2t=1+\tan^2t=1+u^2. Integramos:
\int\Bigl(\tfrac1u+u\Bigr)du =\ln|u|+\tfrac{u^2}{2}+C =\ln|\tan t|+\tfrac12\tan^2t+C.

Passo 6

Para \displaystyle\int_{0}^{\frac\pi3}\tan^2x\operatorname{sec} x\,dx, usamos \tan^2x=\operatorname{sec}^2x-1, então
\int\tan^2x\operatorname{sec} x\,dx =\int(\operatorname{sec}^3x-\operatorname{sec} x)\,dx.
Sabemos que
\int\operatorname{sec} x\,dx=\ln\bigl|\operatorname{sec} x+\tan x\bigr|+C, \quad \int\operatorname{sec}^3x\,dx =\tfrac12\bigl(\operatorname{sec} x\tan x+\ln|\operatorname{sec} x+\tan x|\bigr)+C.
Subtraindo, obtemos
\int(\operatorname{sec}^3x-\operatorname{sec} x)\,dx =\tfrac12\operatorname{sec} x\tan x-\tfrac12\ln\!\bigl|\operatorname{sec} x+\tan x\bigr|+C.
Avaliando de 0 a \pi/3:
em x=\pi/3 tem \operatorname{sec} x=2, \tan x=\sqrt3, então \tfrac12\operatorname{sec} x\tan x=\sqrt3 e \ln|\operatorname{sec} x+\tan x|=\ln(2+\sqrt3).
Em x=0 tudo dá zero. Logo o valor é
\sqrt3-\tfrac12\ln(2+\sqrt3).

Resposta

(a) -\tfrac23\cos^{\tfrac32}x+\tfrac27\cos^{\tfrac72}x+C
(b) \tfrac{16}{105}
(c) \tfrac18\bigl(t-\tfrac{\sin4t}{4}\bigr)+C
(d) \ln|\tan t|+\tfrac12\tan^2t+C
(e) \displaystyle \sqrt3-\tfrac12\ln(2+\sqrt3)

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