Integrate - -sin2x-1 - cos x dx

The correct option is C x sin x+c = ∫sin^2x1+cos xdx = ∫1 cos^2 x1 + cos x dx #160; =∫(1 cos x)(1 cos x)(1+cos x)dx #160; =∫ (1 ...

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First Fundamental Theorem of Calculus

Integrate :- ...

Question

Integrate :-
$\int \frac{{sin}^{2} x}{1 + cos x} d x$

x + cos x + c

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x + sin x + c

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x - sin x + c

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None of these

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I = \int \frac{{sin}^{2} x + sin x}{1 + sin x + cos x} d x

J = \int \frac{{cos}^{2} x + cos x}{1 + sin x + cos x} d x

c

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

\frac{d (I + J)}{d x}

x = \sqrt{2}

= \int \frac{s i n^{2} x + s i n x}{1 + s i n x + c o s x} d x

K = \int \frac{c o s^{2} x + c o s x}{1 + s i n x + c o s x} d x

$y = \frac{s i n x}{x + c o s x}, t h e n \frac{d y}{d x} = ?$

x \in (\frac{π}{4}, \frac{3 π}{4}),

\int \frac{sin x - cos x}{\sqrt{1 - sin 2 x}} e^{sin x}

cos x d x =

I = \int \frac{{sin}^{2} x + sin x}{1 + sin x + cos x} d x

J = \int \frac{{cos}^{2} x + cos x}{1 + sin x + cos x} d x

c

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

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