# ISI2016-DCG-46

1 vote
53 views

Let $I=\int(\sin\:x-\cos\:x)(\sin\:x+\cos\:x)^{3}dx$ and $K$ be a constant of integration. Then the value of $I$ is

1. $(\sin\:x+\cos\:x)^{4}+K$
2. $(\sin\:x+\cos\:x)^{2}+K$
3. $-\frac{1}{4}(\sin\:x+\cos\:x)^{4}+K$
4. None of these
in Calculus
retagged
1
Option C is correct

1 vote

$I = \int \left ( \sin x - \cos x \right )\left (\sin x + \cos x \right )^{3}dx$

$I = \int \left ( \sin x - \cos x \right )\left (\sin x + \cos x \right )\left (\sin x + \cos x \right )^{2}dx$

$I = \int \left (\sin ^{2}x - \cos^{2} x \right )\left (\sin x + \cos x \right )^{2}dx$

$I = - \int \left (\cos^{2} x - \sin ^{2}x \right )\left (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x \right )dx$

$I = - \int \left (\cos 2x \right )\left (1 + \sin 2x \right )dx$

$Let \: 1 + \sin 2x = t$

$Then \: \cos 2x \, dx \: = \: \frac{dt}{2}$

$I = -\int t\: \frac{dt}{2}$

$I = \frac{-\:t^{2}}{4} + K$

$I = \frac{-\: \left ( 1\:+\:\sin 2x \right )^{2}}{4} + K$

$I = \frac{-\: \left ( \sin^2x \:+ \:\cos^2x \:+\: 2\sin x \cos x \right )^{2]}}{4} + K$

$I = \frac{-\: \left ( \sin x \:+\: \cos x \right)^{4]}}{4} + K$

selected by

## Related questions

1
44 views
The Taylor series expansion of $f(x)=\ln(1+x^{2})$ about $x=0$ is $\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n}}{n}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{n+1}$ $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n+1}}{n+1}$
The piecewise linear function for the following graph is $f(x)=\begin{cases} = x,x\leq-2 \\ =4,-2<x<3 \\ = x+1,x\geq 3\end{cases}$ $f(x)=\begin{cases} = x-2,x\leq-2 \\ =4,-2<x<3 \\ = x-1,x\geq 3\end{cases}$ $f(x)=\begin{cases} = 2x,x\leq-2 \\ =x,-2<x<3 \\ = x+1,x\geq 3\end{cases}$ $f(x)=\begin{cases} = 2-x,x\leq-2 \\ =4,-2<x<3 \\ = x+1,x\geq 3\end{cases}$
The general solution of the differential equation $2y{y}'-x=0$ is (assuming $C$ as an arbitrary constant of integration) $x^{2}-y^{2}=C$ $2x^{2}-y^{2}=C$ $2y^{2}-x^{2}=C$ $x^{2}+y^{2}=C$
The general solution of the differential equation $x+y-x{y}'=0$ is (assuming $C$ as an arbitrary constant of integration) $y=x(\log x+C)$ $x=y(\log y+C)$ $y=x(\log y+C)$ $y=y(\log x+C)$