Q114 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

If $I_n=\int \cos ^n x d x$, then $6 I_6-5 I_4=$

  • A
    $-\cos ^5 x \sin ^2 x$
  • B
    $\cos ^6 x \sin ^2 x$
  • C
    $\cos ^3 x \sin ^2 x$
  • Correct answer
    D
    $\cos ^5 x \sin x$

Solution

$$ \begin{aligned} & I_n=\int \cos ^n x d x \\ & =\int \cos ^{n t-1} x \cdot \cos x d x \\ & =\cos ^{n t-1} x \cdot \sin x-\int(n-1) \cos ^{n t-2} x \\ & (-\sin x) \cdot(\sin x) d x \\ & =\cos ^{n t-1} x \sin x+\int \cos ^{n t-2} x \cdot(n-1) \sin ^2 x d x \\ & =\cos ^{n+1} x \sin x+(n-1) \int \cos ^{n t-2} x\left(1-\cos ^2 x\right) d x \\ & =\cos ^{n-1} x \sin x+(n-1) \int \cos ^{n-2} x-(n-1) \\ & \int \cos ^n x d x \\ & =\cos ^{n-1} \chi \sin \chi+(n-1) I_{n-2}-(n-1) I_n \\ & \therefore I_n(1+n-1)=\cos ^{n-1} x \sin x+(n-1) Y_{n t}-2 \\ & \Rightarrow n I_{n t}-(n-1) I_{n-2}=\cos ^{n-1} x \sin x \\ & \end{aligned} $$ Putting $n=6$, we get $$ 6 I_6-5 I_4=\cos ^5 x \sin x $$

Hence, the correct answer is option: 'D'.

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