Bài 110266

Question

Tìm nguyên hàm của hàm số: $f(x)=\cos 5x.\tan x$

score 0 · Answer 1

$F(x)=\int\limits \tan x\cos 5xdx=\int\limits \frac{\cos 5x\sin x}{\cos x}dx= \int\limits \frac{\sin 6x-\sin 4x}{2\cos x}dx $
$F(x)=\frac{1}{2}\int\limits \frac{\sin 6x}{\cos x}dx-\frac{1}{2}\int\limits \frac{\sin 4x}{\cos x}dx $
$F(x)=\frac{1}{2} \int\limits \frac{\sin 2x(3-4\sin^2 2x)}{\cos x}dx- \frac{1}{2} \int\limits \frac{4\sin x\cos x\cos 2x}{\cos x}dx $
$F(x)=\int\limits \sin x(4\cos^2 2x-1)dx-2 \int\limits \sin x\cos 2xdx $
$F(x)=\int\limits \sin x[4(2\cos^2 x-1)^2-1]dx-2\int\sin x(2\cos^2 x-1)dx$
$F(x)=\int\limits \sin x(16\cos^4 x-16\cos^2 x+3)dx-2 \int\limits \sin x(2\cos^2 x-1)dx $
$F(x)=-\int\limits (16\cos^4 x-16\cos^2 x+3)d(\cos x)+2 \int\limits (2\cos^2 x-1)d(\cos x)$
$F(x)=2(\frac{2\cos^3 x}{3}-\cos x )-(\frac{16\cos^5 x}{5}-\frac{16\cos^3 x}{3}+3\cos x )+C$
$F(x)=-\frac{16\cos^5 x}{5}+\frac{20\cos^3 x}{3}-5\cos x+C(ycbt) $

Bài 110266

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Bài 110266

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