How do you integrate $$\int {\arcsin x} \,dx$$ by integration by parts method?

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Vedantu Content Team · Accepted Answer

Hint: We are asked to find the given integral using integration by parts method. First recall the formula for integration by parts method, observe the given term whether it is in form to apply integration by parts method and if not then try to transform the given term into a form so that we can apply integration of parts method.Complete step-by-step answer:We are asked how we will integrate the term $$\int {\arcsin x} \,dx$$ using integration by parts method.First let us know what integration by parts method is. Integration by parts is used when there are two functions such as $$\int {uv\,dx} $$ where $$u$$ and $$v$$ are the functions of $$x$$ , then using integration by parts we get,$$\int {uv\,dx} = u\int {v\,dx - \int {\dfrac{{du}}{{dx}}\left( {\int {v\,dx} } \right)} } \,dx$$ (i)Here, we have the term, $$\arcsin x$$ which can be written as,$$\arcsin x = {\sin ^{ - 1}}x\,$$So, we are to integrate $$\int {{{\sin }^{ - 1}}x\,dx} \,$$ .Let $$I = \int {{{\sin }^{ - 1}}x\,dx} \,$$ (ii)Let $${\sin ^{ - 1}}x = y$$$$ \Rightarrow x = \sin y$$Differentiating we get,$$dx = \cos y\,dy$$Putting the values of $$x$$ and $$dx$$ in equation (ii) we get,$$I = \int {{{\sin }^{ - 1}}\left( {\sin y} \right)\,\cos y\,dy} \,$$$$ \Rightarrow I = \int {y\cos y\,dy} \,$$Here, $$u = y\,$$ and $$v = \cos y$$ . Using the formula of integrating by parts method from equation (i) we have,$$I = y\int {\cos y\,dy} - \int {\left( {\dfrac{d}{{dy}}\left( y \right)} \right)\left( {\int {\cos y} } \right)\,dy} $$$$ \Rightarrow I = y\sin y - \int {\sin y\,dy + C'} $$  $$C'$$ is constant of integration]$$ \Rightarrow I = y\sin y + \cos y + C$$ (iii)  $$C$$ is constant of integration]We have from trigonometric identities,$${\sin ^2}\theta + {\cos ^2}\theta = 1$$$$ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta $$$$ \Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta } $$Therefore, using this $$\cos y\,$$ can be written as, $$\sqrt {1 - {{\sin }^2}y} $$ substituting this in place of $$\cos y\,$$ we get,$$I = y\sin y + \sqrt {1 - {{\sin }^2}y} + C$$Putting $$\sin y = x$$ and $$y = {\sin ^{ - 1}}x$$ in above equation we get,$$I = \left( {{{\sin }^{ - 1}}x} \right)\left( x \right) + \sqrt {1 - {x^2}} + C$$$$ \Rightarrow I = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C$$$$ \Rightarrow I = x\,{\text{arcsin}}\,x + \sqrt {1 - {x^2}} + C$$Therefore, $$\int {\arcsin x\,dx} = x\,{\text{arcsin}}\,x + \sqrt {1 - {x^2}} + C$$So, the correct answer is $$\int {\arcsin x\,dx} = x\,{\text{arcsin}}\,x + \sqrt {1 - {x^2}} + C$$.Note: There are three main functions in trigonometry, these are sine, cosine and tangent. There are three other trigonometric functions which can be written in terms of the main functions, these are cosecant which is inverse of sine, secant which is reciprocal of cosine and cotangent which is reciprocal of tangent. Also, while solving questions related to trigonometry, you should always remember the basic trigonometric identities.

How do you integrate \[\int {\arcsin x} \,dx\] by integration by parts method?