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Rajeev Malhotra 8 years, 1 month ago
{tex}cosA/1-Sin A +Sin A/1- Cos A +1{/tex}
{tex}Cos A(1-Cos A)+Sin A(1-Sin A)+(1-Sin A)(1-Cos A)/(1-Sin A) (1-Cos A){/tex}
{tex}Cos A- Cos^2A+Sin A- Sin^2A+1-SinA- Cos A+Sin ACos A/(1-Sin A)(1-CosA){/tex}
{tex}1-(Cos^2A+Sin ^2A)+SinA CosA/ (1-SinA)(1-CosA){/tex}
{tex}1-1+SinA Cos A/ (1-Sin A)(1-CosA){/tex}
{tex}Sin A CosA/ (1-Sin A)(1-CosA){/tex}
L.H.S = R.H.S
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Sia ? 6 years, 6 months ago
Putting {tex}x = \sin \theta \Rightarrow dx = \cos \theta d\theta {/tex}
{tex}\therefore \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} dx{/tex}
{tex}= \int {{\theta ^2}\cos \theta } d\theta {/tex}
[Applying product rule]
{tex} = {\theta ^2}\sin \theta - \int {2\theta \sin \theta d\theta } {/tex}
{tex}= {\theta ^2}\sin \theta - 2\int {\theta \sin \theta d\theta }{/tex}
[Again applying product rule]
{tex}= {\theta ^2}\sin \theta - 2\left[ {\theta \left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]{/tex}
{tex}= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\int {\cos \theta d\theta } {/tex}
{tex}= {\theta ^2}\sin \theta + 2\theta \cos \theta - 2\sin \theta + c{/tex}
{tex}= x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c{/tex}
Posted by Vikrant Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
- Calculus is used to improve the architecture not only of buildings but also of important infrastructures such as bridges. In Electrical Engineering,
- Calculus (Integration) is used to determine the exact length of power cable needed to connect two substations, which are miles away from each other.
Posted by Shivam Kaushal 8 years, 1 month ago
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Jaya Maheshwari 8 years, 1 month ago

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