Question

`int x sec^(2)(1-x^(2))dx`​

Answer 1

${ \boxed{ \begin{array}{cc} \bf \: \to \: let \\ \\ \rm \: I = \displaystyle \int \: \rm \: x \: {sec}^{2} (1 - {x}^{2} ) \: dx \\ \\ = \displaystyle \int \: \rm \:x \: {sec}^{2} \{ - ( {x}^{2} - 1) \} \: dx \\ \\ = \displaystyle \int \: \rm \:x \: {sec}^{2} ( {x}^{2} - 1) \: dx \\ \\ \: \: \: \: \red{ \{ \because \: sec ( - \theta ) = sec \theta \}} \\ \\ \pink{ \boxed{ \begin{array}{cc} \bf \: substitute \: \\ \rm \: \: u = {x}^{2} - 1 \\ \\ \implies \rm \: \frac{du}{dx} = \frac{d}{dx} ( {x}^{2} - 1) = 2x \\ \\ \implies \rm \:dx = \frac{1}{2x} .du\end{array}}} \\ \\ = \frac{1}{2 \: \cancel x} \displaystyle \int \: \rm \: \cancel x \: {sec}^{2} u \: \: du \\ \\ = \frac{1}{2} \displaystyle \int \: \rm \: {sec}^{2}u \: \: du \\ \\ \pink{ \boxed{ \begin{array}{cc} \sf \: we \: know \: that : \\ \\ \displaystyle \int \: \rm \: {sec}^{2} x \: dx = tan \: x + c\end{array}}} \\ \\ \sf \: apply \: this \: formula \\ \\ = \rm \: \frac{1}{2} \times tan \: u + c \\ \\ \rm = \frac{1}{2} \times tan( {x}^{2} - 1) + c \\ \\ \red{ \{ \because \: \rm \: u = {x}^{2} - 1 \} } \\ \\ \\ \end{array}}}$

$\green{ \small{ \rm \therefore \: \displaystyle \int \: \rm \:x \: {sec}^{2} (1 - {x}^{2} ) \: dx = \frac{1}{2} \times tan( {x}^{2} - 1) + c}}$