Evaluate the indefinite integral: $\int sin 35 t {sec}^{2} (cos 35 t) d t$ $int sin 35t sec^2 (cos 35t)dt$ ?

Question

Evaluate the indefinite integral: $\int sin 35 t {sec}^{2} (cos 35 t) d t$ $int sin 35t sec^2 (cos 35t)dt$ ?

1 Answer

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Answer 1 · 2017-01-24T00:59:16

$- \frac{1}{35} tan (cos 35 t) + C$ $-1/35tan(cos35t)+C$

Explanation:

$\int sin 35 t {sec}^{2} (cos 35 t) d t$ $intsin35tsec^2(cos35t)dt$

Use the substitution $u = cos 35 t \Rightarrow d u = - 35 sin 35 t$ $u=cos35t=>du=-35sin35t$ .

$= - \frac{1}{35} \int {sec}^{2} (cos 35 t) (- 35 sin 35 t) d t$ $=-1/35intsec^2(cos35t)(-35sin35t)dt$

$= - \frac{1}{35} \int {sec}^{2} (u) d u$ $=-1/35intsec^2(u)du$

This is a common integral:

$= - \frac{1}{35} tan (u) + C$ $=-1/35tan(u)+C$

$= - \frac{1}{35} tan (cos 35 t) + C$ $=-1/35tan(cos35t)+C$

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Evaluate the indefinite integral: $\int sin 35 t {sec}^{2} (cos 35 t) d t$ $int sin 35t sec^2 (cos 35t)dt$ ?

1 Answer

Explanation:

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Evaluate the indefinite integral: $\int sin 35 t {sec}^{2} (cos 35 t) d t$ $int sin 35t sec^2 (cos 35t)dt$ ?