Integral of sinx+sinx tan^2x/sec^2x
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Integral of sinx+sinx tan^2x/sec^2x

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
sin^3(x) = sin(x)sin²(x) = sin(x)(1 - cos²(x)).∫ sin(x) dx + ∫ sin(x)(1 - cos²(x)) dx = - cos(x) - x + cos^3(x)/3 + C.For the second integral, use the substitution u = cos(x), du = -sin(x) dx.-Split the integral in two.......
Your integrand is not at all clear. As typed (perhaps there are missing parentheses) your function is

sin(x) + sin(x)tan²(x)/sec²(x) = sin(x) + sin(x)tan²(x)cos²(x) = sin(x) + sin^3(x).

The anti-derivative of the sine is the negative of the cosine. As for the cubed term, use

sin^3(x) = sin(x)sin²(x) = sin(x)(1 - cos²(x)).

∫ sin(x) dx + ∫ sin(x)(1 - cos²(x)) dx = - cos(x) - x + cos^3(x)/3 + C.

For the second integral, use the substitution u = cos(x), du = -sin(x) dx.

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Split the integral in two. sinx tan²x/sec²x = sin³x/cos⁴x. Let u = cos⁴x, du = –4sin³x dx. You should be able to take it from there.
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