Integral Trigonometri : Soal dan Pembahasan

Berikut ini adalah kumpulan soal dan pembahasan mengenai integral trigonometri. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Nilai k antara 0 dan \pi yang membuat \displaystyle\int_0^k\sin^2x\cos x\ dx menjadi maksimum adalah....

$\begin{array}{rl}\underset{0}{\overset{k}{\int }}{\sin }^{2}x\cos xdx& =\underset{0}{\overset{k}{\int }}{\sin }^{2}xd(\sin x)\\ & ={[{\displaystyle \frac{1}{3}}{\sin }^{3}x]}_0^k\\ & ={\displaystyle \frac{1}{3}}{\sin }^{3}k-{\displaystyle \frac{1}{3}}{\sin }^3(0)\\ & ={\displaystyle \frac{1}{3}}{\sin }^{3}k\end{array}$

Agar menjadi maksimum, maka \sin k=1 atau k=\dfrac{\pi}2.

Jadi, k=\dfrac{\pi}2.

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Hasil dari \displaystyle\int\dfrac{1+\dfrac1{\csc^2x-1}}{4\tan^2x+2\tan x+3}\ dx adalah ....

http://www.learncy.net/problem/177/

$$\begin{array}{rl}{\displaystyle \int {\displaystyle \frac{1+{\displaystyle \frac{1}{{\csc }^{2}x-1}}}{4{\tan }^{2}x+2\tan x+3}}dx}& ={\displaystyle \int {\displaystyle \frac{1+{\displaystyle \frac{1}{{\cot }^{2}x}}}{4{\tan }^{2}x+2\tan x+3}}\cdot {\displaystyle \frac{4}{4}}dx}\\ & ={\displaystyle \int {\displaystyle \frac{4(1+{\tan }^{2}x)}{16{\tan }^{2}x+8\tan x+12}}dx}\\ & ={\displaystyle \int {\displaystyle \frac{4{\sec }^{2}x}{16{\tan }^{2}x+8\tan x+1+11}}dx}\\ & ={\displaystyle \int {\displaystyle \frac{4{\sec }^{2}x}{(4\tan x+1{)}^2+11}}dx}\\ & ={\displaystyle \int {\displaystyle \frac{1}{(4\tan x+1{)}^2+11}}d(4\tan x+1)}\\ & ={\displaystyle \frac{1}{\sqrt{11}}}{\tan }^{-1}\left({\displaystyle \frac{4\tan x+1}{\sqrt{11}}}\right)+C\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{\sqrt{11}}{11}}{\tan }^{-1}({\displaystyle \frac{\sqrt{11}}{11}}(4\tan x+1))+C}}}}\end{array}$$

Jadi, \displaystyle\int\dfrac{1+\dfrac1{\csc^2x-1}}{4\tan^2x+2\tan x+3}\ dx=\dfrac{\sqrt{11}}{11}\tan^{-1}\left(\dfrac{\sqrt{11}}{11}(4\tan x+1)\right)+C.
JAWAB: C

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No. 3

\displaystyle\int x\cos2x\ dx=

SUMBER

Misal

$$\begin{array}{rl}u& =x\\ du& =dx\end{array}$$

$$\begin{array}{rl}dv& =\cos 2x\\ v& ={\displaystyle \frac{1}{2}}\sin 2x\end{array}$$

$$\begin{array}{rl}{\displaystyle \int udv}& =uv-{\displaystyle \int vdu}\\ {\displaystyle \int x\cos 2xdx}& =x({\displaystyle \frac{1}{2}}\sin 2x)-{\displaystyle \int {\displaystyle \frac{1}{2}}\sin 2xdx}\\ & ={\displaystyle \frac{1}{2}}x\sin 2x+{\displaystyle \frac{1}{4}}\cos 2x+C\end{array}$$

Jadi, \displaystyle\int x\cos2x\ dx=\dfrac12x\sin2x+\dfrac14\cos2x+C.

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