Question

Calcule as integrais por substituição
A) ∫12t^6cos(3t^7+1)dt

b) ∫3x²+x/4x³+2x²-6

C) ∫3x²(4x³-3)^5 dx

Answer 1

A)

$\displaystyle\int12t^{6}cos(3t^{7}+1)dt$

Escolheremos v sendo a função que está dentro do cosseno:

$v=3t^{7}+1$

Derivando:

$dv=3\cdot7t^{7-1}dt\\\\dv=21t^{6}dt\\\\(\frac{12}{21})dv=(\frac{12}{21})21t^{6}dt\\\\\boxed{\boxed{\frac{4}{7}dv=12t^{6}dt}}$

Introduzindo v e dv na integral:

$\displaystyle\int12t^{6}cos(3t^{7}+1)dt=\int cos(3t^{7}+1)\cdot12t^{6}dt\\\\\\\int12t^{6}cos(3t^{7}+1)dt=\int cos(v)\cdot\dfrac{4}{7}dv\\\\\\\int12t^{6}cos(3t^{7}+1)dt=\dfrac{4}{7}\int cos(v)dv\\\\\\\int12t^{6}cos(3t^{7}+1)dt=\dfrac{4}{7}sen(v)+constante\\\\\\\boxed{\boxed{\int12t^{6}cos(3t^{7}+1)dt=\dfrac{4}{7}sen(3t^{7}+1)+constante}}$
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B)

$\displaystyle\int\dfrac{3x^{2}+x}{4x^{3}+2x^{2}-6}dx$

Chamando o denominador da fração de t:

$t=4x^{3}+2x^{2}-6\\\\dt=(4\cdot3x^{3-1}+2\cdot2x^{2-1})dx\\\\dt=(12x^{2}+4x)dx\\\\dt=4(3x^{2}+x)dx\\\\\boxed{\boxed{\dfrac{dt}{4}=(3x^{2}+x)dx}}$

Então:

$\displaystyle\int\dfrac{3x^{2}+x}{4x^{3}+2x^{2}-6}dx=\int\dfrac{(\frac{1}{4})}{t}dt\\\\\\\int\dfrac{3x^{2}+x}{4x^{3}+2x^{2}-6}dx=\dfrac{1}{4}\int\dfrac{1}{t}dt\\\\\\\int\dfrac{3x^{2}+x}{4x^{3}+2x^{2}-6}dx=\dfrac{1}{4}ln|t|+constante\\\\\\\boxed{\boxed{\int\dfrac{3x^{2}+x}{4x^{3}+2x^{2}-6}dx=\dfrac{1}{4}ln|4x^{3}+2x^{2}-6|+constante}}$
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C)

$\displaystyle\int3x^{2}(4x^{3}-3)^{5}dx$

Vamos definir 'a' como sendo 4x³ - 3:

$a=4x^{3}-3\\\\da=4\cdot3x^{3-1}dx\\\\da=12x^{2}dx\\\\da=4\cdot3x^{2}dx\\\\\boxed{\boxed{\dfrac{1}{4}da=3x^{2}dx}}$

Trocando 4x³ - 3 por 'a' e 3x²dx por (1/4)da na integral:

$\displaystyle\int3x^{2}(4x^{3}-3)^{5}dx=\int(4x^{3}-3)^{5}\cdot3x^{2}dx\\\\\\\int3x^{2}(4x^{3}-3)^{5}dx=\int a^{5}\left(\frac{1}{4}\right)da\\\\\\\int3x^{2}(4x^{3}-3)^{5}dx=\dfrac{1}{4}\int a^{5}da\\\\\\\int3x^{2}(4x^{3}-3)^{5}dx=\dfrac{1}{4}\cdot\dfrac{a^{6}}{6}+constante\\\\\\\boxed{\boxed{\int3x^{2}(4x^{3}-3)^{5}dx=\dfrac{1}{24}(4x^{3}-3)^{6}+constante}}$