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....

$\eqalign{
\intop_0^k\sin^2x\cos x\ dx&=\intop_0^k\sin^2x\ d(\sin x)\\
&=\left[\dfrac13\sin^3x\right]_0^k\\
&=\dfrac13\sin^3k-\dfrac13\sin^3(0)\\
&=\dfrac13\sin^3k
}$

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{aligned}
\displaystyle\int\dfrac{1+\dfrac1{\csc^2x-1}}{4\tan^2x+2\tan x+3}\ dx&=\displaystyle\int\dfrac{1+\dfrac1{\cot^2x}}{4\tan^2x+2\tan x+3}\cdot\dfrac44\ dx\\[8pt]
&=\displaystyle\int\dfrac{4\left(1+\tan^2x\right)}{16\tan^2x+8\tan x+12}\ dx\\[8pt]
&=\displaystyle\int\dfrac{4\sec^2x}{16\tan^2x+8\tan x+1+11}\ dx\\[8pt]
&=\displaystyle\int\dfrac{4\sec^2x}{(4\tan x+1)^2+11}\ dx\\[8pt]
&=\displaystyle\int\dfrac1{(4\tan x+1)^2+11}\ d(4\tan x+1)\\[8pt]
&=\dfrac1{\sqrt{11}}\tan^{-1}\left(\dfrac{4\tan x+1}{\sqrt{11}}\right)+C\\
&=\boxed{\boxed{\dfrac{\sqrt{11}}{11}\tan^{-1}\left(\dfrac{\sqrt{11}}{11}(4\tan x+1)\right)+C}}
\end{aligned}

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{aligned}
u&=x\\
du&=dx
\end{aligned}

\begin{aligned}
dv&=\cos2x\\
v&=\dfrac12\sin2x
\end{aligned}

\begin{aligned}
\displaystyle\int u\ dv&=uv-\displaystyle\int v\ du\\
\displaystyle\int x\cos2x\ dx&=x\left(\dfrac12\sin2x\right)-\displaystyle\int\dfrac12\sin2x\ dx\\
&=\dfrac12x\sin2x+\dfrac14\cos2x+C
\end{aligned}

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

3