For n ge 2,if I_n = int (sin x + cos x)^n dx, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I_n = \int (\sin x + \cos x)^n dx$. Using integration by parts,let $u = (\sin x + \cos x)^{n-1}$ and $dv = (\sin x + \cos x) dx$. Then $du =

Vedclass · Accepted Answer

$(\sin x + \cos x)^{n-1}(\sin x - \cos x) + C$

Vedclass · Answer

(C) Let $I_n = \int (\sin x + \cos x)^n dx$. Using integration by parts,let $u = (\sin x + \cos x)^{n-1}$ and $dv = (\sin x + \cos x) dx$. Then $du = (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x) dx$ and $v = (-\cos x + \sin x) = (\sin x - \cos x)$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - \int (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x)(\sin x - \cos x) dx$. Since $(\cos x - \sin x)(\sin x - \cos x) = -(\sin x - \cos x)^2 = -(1 - \sin 2x) = \sin 2x - 1$. Also,$(\sin x + \cos x)^2 = 1 + \sin 2x$,so $\sin 2x = (\sin x + \cos x)^2 - 1$. Thus,$\sin 2x - 1 = (\sin x + \cos x)^2 - 2$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) \int (\sin x + \cos x)^{n-2} ((\sin x + \cos x)^2 - 2) dx$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) I_n + 2(n-1) I_{n-2}$. $I_n + (n-1) I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$. $n I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$. $n I_n - 2(n-1) I_{n-2} = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + C$.

For $n \ge 2$,if $I_n = \int (\sin x + \cos x)^n dx$,then $nI_n - 2(n-1)I_{n-2} = $

Similar Questions

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$ then $\frac{m-l}{r}=$

$\int \frac{x^4+1}{x^6+1} \, dx =$

If $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} \,d x$ and $J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} \,d x$ then for any arbitrary constant $c$, the value of $J-I$ equals

If $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx = \sin^{-1}(f(x)) + C$,(where $C$ is the constant of integration),then $f(2)$ has the value:

If $\int \frac{x-\sin x}{1+\cos x} dx = x \tan \left(\frac{x}{2}\right) + p \log \left|\sec \left(\frac{x}{2}\right)\right| + C$,then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module