Resolver a integral abaixo:
\int_{0}^{\pi / 4} \frac{x^{3}}{\sqrt{x^{2}+25}} d x. Dica: Use a substituição x=5 \operatorname{tg} \theta

Passo 1

O enunciado pede para calcular
\int_{0}^{\pi/4} \frac{x^{3}}{\sqrt{x^{2}+25}}\,dx
e sugere usar a substituição trigonométrica \,x=5\tan\theta.
Esse tipo de substituição é útil quando aparece \sqrt{x^{2}+a^{2}}, pois
x^{2}+25=25\tan^{2}\theta+25=25\operatorname{sec}^{2}\theta.

Passo 2

Definimos
x=5\tan\theta, \quad dx=5\operatorname{sec}^{2}\theta\,d\theta, \quad \sqrt{x^{2}+25}=5\operatorname{sec}\theta.
Quando x=0 temos \theta=0, e quando x=\pi/4 temos \theta=\arctan\bigl(\tfrac{\pi}{20}\bigr).

Passo 3

Substituindo na integral, o numerador x^{3}=(5\tan\theta)^{3}=125\tan^{3}\theta e o denominador \sqrt{x^{2}+25}=5\operatorname{sec}\theta. Assim
\frac{x^{3}}{\sqrt{x^{2}+25}}\,dx = \frac{125\tan^{3}\theta}{5\operatorname{sec}\theta}\,\bigl(5\operatorname{sec}^{2}\theta\,d\theta\bigr) =125\,\tan^{3}\theta\cos\theta\,\operatorname{sec}^{2}\theta\,d\theta.
Como \cos\theta\,\operatorname{sec}^{2}\theta=\operatorname{sec}\theta, fica
125\,\tan^{3}\theta\,\operatorname{sec}\theta =125\,\frac{\sin^{3}\theta}{\cos^{4}\theta} =125\bigl(\operatorname{sec}^{3}\theta\tan\theta-\operatorname{sec}^{2}\theta\tan\theta\bigr).

Passo 4

Agora integramos em \theta:
\int 125\bigl(\operatorname{sec}^{3}\theta\tan\theta-\operatorname{sec}^{2}\theta\tan\theta\bigr)\,d\theta =125\Bigl(\frac{\operatorname{sec}^{3}\theta}{3}-\frac{\operatorname{sec}^{2}\theta}{2}\Bigr)+C.

Passo 5

Voltando para x, usamos \operatorname{sec}\theta=\frac{\sqrt{x^{2}+25}}{5}. Então
\frac{\operatorname{sec}^{3}\theta}{3} =\frac{1}{3}\Bigl(\frac{\sqrt{x^{2}+25}}{5}\Bigr)^{3} =\frac{(x^{2}+25)^{3/2}}{375}, \quad \frac{\operatorname{sec}^{2}\theta}{2} =\frac{1}{2}\Bigl(\frac{x^{2}+25}{25}\Bigr) =\frac{x^{2}+25}{50}.
Multiplicando por 125, obtemos a antiderivada
F(x) =\frac{1}{3}(x^{2}+25)^{3/2} -\frac{5}{2}(x^{2}+25)+C.

Passo 6

Avaliamos de 0 a \pi/4:
F\bigl(\tfrac{\pi}{4}\bigr) =\frac{1}{3}\Bigl(\bigl(\tfrac{\pi}{4}\bigr)^{2}+25\Bigr)^{3/2} -\frac{5}{2}\Bigl(\bigl(\tfrac{\pi}{4}\bigr)^{2}+25\Bigr),
F(0) =\frac{1}{3}(25)^{3/2} -\frac{5}{2}\cdot25 =\frac{125}{3}-\frac{125}{2} =-\frac{125}{6}.
Logo
\int_{0}^{\pi/4}\frac{x^{3}}{\sqrt{x^{2}+25}}\,dx =F\bigl(\tfrac{\pi}{4}\bigr)-F(0) =\frac{1}{3}\Bigl(\frac{\pi^{2}}{16}+25\Bigr)^{3/2} -\frac{5}{2}\Bigl(\frac{\pi^{2}}{16}+25\Bigr) +\frac{125}{6}.

Resposta

\int_{0}^{\pi/4}\frac{x^{3}}{\sqrt{x^{2}+25}}\,dx =\frac{1}{3}\Bigl(\frac{\pi^{2}}{16}+25\Bigr)^{3/2} -\frac{5}{2}\Bigl(\frac{\pi^{2}}{16}+25\Bigr) +\frac{125}{6}.

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