Solution: Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{d} \mathrm{t} \\ \Rightarrow...

Solution: Let $x-\cos x=t$ $\begin{array}{l} \Rightarrow d(x-\cos x)=d t \\ \Rightarrow(1+\sin x) d x=d t \end{array}$ $\therefore$ By substituting $\mathrm{t}$ and dt in given equation we obtain...

Solution: Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{dt} \\ \Rightarrow...

Solution: Let $\cot x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cot x)=d t \\ \Rightarrow-\operatorname{cosec}^{2} x \cdot d x=d t \\ \Rightarrow d x=\frac{-d t}{\csc ^{2} x} \end{array}$...

Solution: Let $1+\mathrm{e}^{\mathrm{x}}=\mathrm{t}$ $\Rightarrow d\left(1+e^{x}\right)=d t$ $\Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$ $\therefore$ By substituting $t$ and $dt$...

Solution: Let $\cos x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cos x)=d t \\ \Rightarrow-\sin x d x=d t \\ \Rightarrow d x=\frac{-d t}{\sin x} \end{array}$ $\therefore$ On substituting...

Solution: Let $1+e^{x}=t$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(1+\mathrm{e}^{\mathrm{x}}\right)=\mathrm{dt} \\ \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}=\mathrm{d}...

Solution: Let $1+v x=t$ $\begin{array}{l} \Rightarrow d(1+v x)=d t \\ \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t \end{array}$ $\therefore$ On substituting...

Solution: Let $\log \left(1+\frac{1}{\mathrm{x}}\right)=\mathrm{t}$ $\begin{array}{l} \Rightarrow \operatorname{d}\left(\log \left(1+\frac{1}{\mathrm{x}}\right)\right)=\mathrm{dt} \\ \Rightarrow...

Solution: Let $\log x=t$ $\begin{array}{l} \Rightarrow d(\log x)=d t \\ \Rightarrow \frac{1}{x} d x=d t \end{array}$ By substituting $\mathrm{t}$ and $dt$ in above equation we obtain...