Question

Let $I=\int \frac{{\sin }^{2}x+\sin x}{1+\sin x+\cos x}dx$, $J=\int \frac{{\cos }^{2}x+\cos x}{1+\sin x+\cos x}dx$ and $c$ is the constant of integration.

$$\begin{array}{cc}\text{Function}& \text{Integral}\\ \begin{array}{c}\text{(a) I}\end{array}& \text{(p)}\frac{1}{2}(x-\sin x-\cos x)+c\\ \begin{array}{c}\text{(b) J}\end{array}& \text{(q)}\frac{1}{2}(x+\sin x+\cos x)+c\\ \begin{array}{c}\text{(c) I + J}\end{array}& \text{(r)}x+c\\ \begin{array}{c}\text{(d) I - J}\end{array}& \text{(s)}c-\cos x-\sin x\\ \begin{array}{}\end{array}& \text{(t)}c+\cos x+\sin x\\ \begin{array}{}\end{array}& \text{(u)}-\frac{1}{2}(x+\sin x+\cos x+c)\end{array}$$

Then the value of $\frac{d(I+J)}{dx}$ at $x=\sqrt{2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$I+J=\int \frac{{\sin }^{2}x+{\cos }^{2}x+\sin x+\cos x}{1+\sin x+\cos x}dx$

$\Rightarrow I+J=x+c_1\cdots \left(1\right)$

$I-J=\int \frac{{\sin }^{2}x+\sin x-{\cos }^{2}x-\cos x}{\sin x+\cos x+1}\Rightarrow I-J=\frac{(\sin x-\cos x)(1+\sin x+\cos x)}{1+\sin x+\cos x}$

$\Rightarrow I-J=-\cos x-\sin x+c_2\cdots \left(2\right)$

$\left(1\right)+\left(2\right)$ gives
$I=\frac{1}{2}(x-\sin x-\cos x)+c$

$\left(1\right)-\left(2\right)$ gives
$J=\frac{1}{2}(x+\sin x+\cos x)+c$