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TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

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Question : 74 of 160

Marks: +1, -0

I_n=\int x^n \sin x\,dx

I_6-360\;I_2=f(x)\cos\,x+g(x)\sin\,x

f(1)+g(1)=

$-85$
0
$-53$
$75$

Solution:

We have,

I_{n}=\int x^{n} \sin x\,dx

\therefore I_{6}=\int x^{6} \sin x\,dx

=x^{6}(-\cos x)-\int (-\cos x)\cdot(6 x^{5})\,dx

=-x^{6}\cos x+6\int x^{5}\cos x\,dx

=-x^{6}\cos x+6\left[x^{5}\sin x-\int \sin x\cdot(5 x^{4})\,dx\right]

=-x^{6}\cos x+6 x^{5}\sin x-30\int x^{4}\sin x\,dx

=-x^{6}\cos x+6 x^{5}\sin x-30\left[x^{4}(-\cos x)-\int (-\cos x)(4 x^{3})\,dx\right]

=-x^{6}\cos x+6 x^{5}\sin x+30 x^{4}\cos x\; -120\int x^{3}\cos x\,dx

=-x^{6}\cos x+6 x^{5}\sin x+30 x^{4}\cos x-120\left[x^{3}\sin x-\int \sin x(3 x^{2})\,dx\right]

=-x^{6}\cos x+6 x^{5}\sin x+30 x^{4}\cos x-120 x^{3}\sin x +360\int x^{2}\sin x\,dx

=-x^{6}\cos x+6 x^{5}\sin x+30 x^{4}\cos x

-120 x^{3}\sin x+360 I_{2}

\therefore I_{6}-360 I_{2}=(-x^{6}+30 x^{4})\cos x

+(6 x^{5}-120 x^{3})\sin x

\therefore \; f(x)=-x^{6}+30 x^{4}

and

g(x)=6 x^{5}-120 x^{3}

\therefore \; f(1)+g(1)=-1+30+6-120=-85

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