Q. 74 - Calculus

Question

Prove $$\begin{gathered} \int \sin x dx=-\cos x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(b\right) \int \cos x dx=\sin x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(c\right)\int se{c}^2 x dx=\tan x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{d}\right) \int {\mathrm{cosec}}^2 \mathrm{x} \mathrm{dx}=-\mathrm{cotx} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{e}\right)\int {\sec }^{} \mathrm{xtan} \mathrm{x} \mathrm{dx}=\sec \mathrm{x} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{f}\right)\int \mathrm{cosec} \mathrm{xcotx} \mathrm{dx}=-\mathrm{cosec} \mathrm{x} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \end{gathered}$$

Step-by-Step Solution

Answer

Hence proved

Step 1. To prove

$$\begin{gathered} \left(a\right) \int \sin x dx=-\cos x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(b\right) \int \cos x dx=\sin x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(c\right)\int se{c}^2 x dx=\tan x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{d}\right) \int {\mathrm{cosec}}^2 \mathrm{x} \mathrm{dx}=-\mathrm{cotx} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{e}\right)\int {\sec }^{} \mathrm{xtan} \mathrm{x} \mathrm{dx}=\sec \mathrm{x} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \left(\mathrm{f}\right)\int \mathrm{cosec} \mathrm{xcotx} \mathrm{dx}=-\mathrm{cosec} \mathrm{x} \mathrm{dx}+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \end{gathered}$$

Part(a) Step 2. Proving part a

$$\begin{gathered} \left(a\right)\int \sin x dx=-\cos x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\left(-\cos x\right)=\sin x \\ so -\cos x \mathrm{is} \mathrm{antiderivative} \mathrm{of} \sin \mathrm{x} \\ \mathrm{hence} \mathrm{proved} \int \sin \mathrm{x} \mathrm{dx}=-\cos \mathrm{x} \mathrm{dx}+\mathrm{C} \end{gathered}$$

Part (b) Step 3. Proving

$$\begin{gathered} \left(b\right)\int \cos x dx=\sin x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\left(\sin x\right)=\cos x \\ so -\cos x \mathrm{is} \mathrm{antiderivative} \mathrm{of} \cos \mathrm{x} \\ \mathrm{hence} \mathrm{proved} \int \cos x\mathrm{dx}=\sin \mathrm{x} \mathrm{dx}+\mathrm{C} \end{gathered}$$

Part (c) Step 4. Proving

$$\begin{gathered} \left(c\right)\int se{c}^2 x dx=\tan x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\left(se{c}^2 x\right)=\tan x \\ so \left(se{c}^2 x\right) \mathrm{is} \mathrm{antiderivative} \mathrm{of} \tan x \\ \mathrm{hence} \mathrm{proved} \int se{c}^2 x dx=\tan x dx+\mathrm{C} \end{gathered}$$

Part (d) Step 5. Proving

$$\begin{gathered} \left(d\right)\int cose{c}^2 x dx=-cot x dx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\left(cose{c}^2 x\right)=cot x \\ so \left(cose{c}^2 x\right) \mathrm{is} \mathrm{antiderivative} \mathrm{of} cot x \\ \mathrm{hence} \mathrm{proved} \int {\mathrm{cosec}}^2 \mathrm{x} \mathrm{dx}=-\cot \mathrm{x} \mathrm{dx}+\mathrm{C} \end{gathered}$$

Part(e) Step 6. Proving

$$\begin{gathered} \left(e\right)\int sec x \tan x dx=sec xdx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\left(sec x \tan x \right)=sec x \\ so \left(sec x \tan x\right) \mathrm{is} \mathrm{antiderivative} \mathrm{of} sec x \\ \mathrm{hence} \mathrm{proved} \int {\mathrm{cosec}}^2 \mathrm{x} \mathrm{dx}=-\cot \mathrm{x} \mathrm{dx}+\mathrm{C} \end{gathered}$$

Part (f)Step 7. Proving

$$\begin{gathered} \left(f\right)\int \cos ec x cot xdx=-cot xdx+\mathrm{C} \mathrm{using} \mathrm{differentiation} \mathrm{formulas} \\ \frac{d}{dx}\cos ec x cot x=-cot x \\ so\cos ec x cot x \mathrm{is} \mathrm{antiderivative} \mathrm{of}-cot x \\ \mathrm{hence} \mathrm{proved} \int \mathrm{cosec} \mathrm{x} \cot \mathrm{xdx}=-\cot \mathrm{xdx}+\mathrm{C} \end{gathered}$$