If ∫ cos n x sin x dx =-( cos 6 x /6)+ C, then n =

Question

If ∫ cos n x sin x dx =-( cos 6 x /6)+ C, then n = (A) 0 (B) 1 (C) 2 (D) 5

Tardigrade Admin · Accepted Answer

Let I =∫ cos n x sin x d x Put cos x = t - sin x dx = dt ∴ I =-∫ t n dt =-( t n +1/ n +1)+ C =-( cos n +1 x / n +1)+ C =-( cos 6 x /6)+ C ∴ n +1=6 or n =5

Q. If $\int cos^{n} x sin x d x = - \frac{c o s ^{6} x}{6} + C$ , then $n =$

0

1

2

5

Let $I = \int cos^{n} x sin x d x$
Put $cos x = t$
$- sin x d x = d t$
$∴ I = - \int t^{n} d t = - \frac{t ^{n + 1}}{n + 1} + C$
$= - \frac{c o s ^{n + 1} x}{n + 1} + C$
$= - \frac{c o s ^{6} x}{6} + C$
$∴ n + 1 = 6$ or $n = 5$

Q. If ∫cosnxsinxdx=−6cos6x​+C, then n=

0

1

2

5

Let I=∫cosnxsinxdx Put cosx=t −sinxdx=dt ∴I=−∫tndt=−n+1tn+1​+C =−n+1cosn+1x​+C =−6cos6x​+C ∴n+1=6 or n=5

Q. If $\int cos^{n} x sin x d x = - \frac{c o s ^{6} x}{6} + C$ , then $n =$

Let $I = \int cos^{n} x sin x d x$
Put $cos x = t$
$- sin x d x = d t$
$∴ I = - \int t^{n} d t = - \frac{t ^{n + 1}}{n + 1} + C$
$= - \frac{c o s ^{n + 1} x}{n + 1} + C$
$= - \frac{c o s ^{6} x}{6} + C$
$∴ n + 1 = 6$ or $n = 5$