Question

Integral de
${tg}^{2} ( \frac{1}{3} x) {sec}^{2} ( \frac{1}{3} x)$
Resposta:
${tg}^{3} ( \frac{1}{3} x) + c$
​

Answer 1

Explicação passo-a-passo:

Olha na tabela, a integral de Sec^2(1/3 x) = Tan (1/3 x)

Por isso a resposta ficou elevada ao cubo

Tan^2 (1/3 x) Tan (1/3 x)

Tan^3 (1/3 x)

Answer 2

Olá,

Temos a integral:

$\tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx$

Vamos resolver por substituição:

$\tt \: u = tg( \frac{1}{3} x) \: \to \: du = {sec}^{2} ( \frac{1}{3} x). \: ( \frac{1}{3} ) \: dx \\ \tt \: dx = \frac{3 \: du}{ {sec}^{2}( \frac{1}{3} x)}$

Substituindo na integral:

$\tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = \int {u}^{2} {sec}^{2} ( \frac{1}{3} x) \frac{3 \: du}{ {sec}^{2} ( \frac{1}{3} x)} \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = 3 \int {u}^{2} \cancel{ {sen}^{2}( \frac{1}{3} x) } \frac{du}{ \cancel{ {sec}^{2}( \frac{1}{3} x)}} \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = 3 \int {u}^{2} du \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = 3 \frac{ {u}^{2 + 1} }{2 + 1} + c \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = 3 \frac{ {u}^{3} }{3} + c \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = \cancel{3} \frac{ {u}^{3} }{ \cancel{3}} + c \\ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = {u}^{3} + c$

Substituindo $\tt \: u = {tg}( \frac{1}{3} x)$

Temos:

$\boxed{ \tt \int {tg}^{2} ( \frac{1}{3} x) \: {sec}^{2} ( \frac{1}{3} x)dx = {tg}^{3} ( \frac{1}{3} x) + c}$