Soal EBTANAS Matematika SMA IPA 1990
Tentukan \( \int (x^2+1) \cos x \ dx = \cdots \ ? \)
- \( x^2 \sin x + 2x \cos x + C \)
- \( (x^2-1) \sin x + 2x \cos x + C \)
- \( (x^2+3) \sin x - 2x \cos x + C \)
- \( 2x^2 \cos x + 2x^2 \sin x + C \)
- \( 2x \sin x - (x^2-1) \cos x + C \)
Pembahasan:
Kita bisa selesaikan soal ini menggunakan teknik integral parsial. Misalkan \( u = x^2 + 1 \) dan \( dv = \cos x \ dx \) sehingga \( \int (x^2+1) \cos x \ dx = \int u \ dv \) dan kita peroleh hasil berikut:
\begin{aligned} u = x^2+1 \Leftrightarrow du &= 2x \ dx \\[8pt] dv = \cos x \ dx \Leftrightarrow v &= \int \cos x \ dx \\[8pt] &= \sin x \end{aligned}
Dari hasil di atas, kita peroleh:
\begin{aligned} \int u \ dv &= uv-\int v \ du \\[8pt] \int (x^2+1) \cos x \ dx &= (x^2+1) \sin x - \int \sin x \cdot 2x dx \\[8pt] &= (x^2+1)\sin x - 2 \int x \sin x \ dx \\[8pt] &= (x^2+1) \sin x - 2 \left( -x \cos x - \int -\cos x \ dx \right) \\[8pt] &= x^2 \sin x + \sin x + 2x \cos x - 2 \sin x + C \\[8pt] &= x^2 \sin x - \sin x + 2x \cos x + C \\[8pt] &= (x^2-1) \sin x + 2x \cos x + C \end{aligned}
Jawaban B.