Tahun berapa sekarang…?

Mungkin itu bisa dijawab dengan mengerjakan soal berikut…

hitunglah \frac{1}{\int _0^{\frac{\pi}{2}} \cos ^{2007}x \cdot \sin {2009 x} dx}

 

Misalkan

I=\int _0^{\frac{\pi}{2}} \cos ^{2007}x \cdot \sin {2009 x} dx

Kita tahu bahwa \sin{(a+b)}=\sin{a}\cos{b}+\cos{a}\sin{b}

Maka kita bisa ubah \sin{2009x} = \sin{(x+2008x)} = \sin{x}\cos{2008x} + \cos{x}\sin{2008x}

Maka,,

I =\int _0^{\frac{\pi}{2}} [\cos ^{2007}x \cdot \sin{x}\cos{2008x}+\cos^{2008}x\cdot\sin{2008x}] dx

I =\int _0^{\frac{\pi}{2}} [\cos{2008x}\cos^{2007}x \cdot \sin{x}+\cos^{2008}x \cdot \sin{2008x}] dx

I = -\frac{1}{2008}\int _0^{\frac{\pi}{2}} \cos{2008x}  d(\cos^{2008}x )+ \cos^{2008}x d(\cos{2008x})

Kita tau bahwa d(uv) = v du+ u dv

Maka,,

I = -\frac{1}{2008}\int _0^{\frac{\pi}{2}} d(\cos{2008x} \cdot \cos^{2008}x)

I = -\frac{1}{2008}[ cos{2008x} \cdot \cos ^{2008}x]_0^{\frac{\pi}{2}}

I =\frac{1}{2008}

dan \frac{1}{I} =2008

Maka solusi dari soal di atas adalah 2008

About mhar_teens

I'm only an ordinary boy... still confused about my future,, I only know that God has a great plan for me View all posts by mhar_teens

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