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Rishika Garg 8 years, 1 month ago

First of all complete your ncert along with examples it took me about 6 hrs to complete it then move on to Rd since we don't have much time so I am thinking of doing only examples ......ps. I am not clear with the properties of definite integration Hope it helps?
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Yugantar Borah 8 years, 1 month ago

3cotx - CPT౩^/ 1 - 3cot २^ 3^ : 3cube. & २^ : square
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Nikhil Verma 8 years, 1 month ago

integration with respect to dy or dx.we don't know this question is incomplete
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Nikhil Verma 8 years, 1 month ago

∫√(tan x) dx Let tan x = t2 ⇒ sec2 x dx = 2t dt ⇒ dx = [2t / (1 + t4)]dt ⇒ Integral ∫ 2t2 / (1 + t4) dt ⇒ ∫[(t2 + 1) + (t2 - 1)] / (1 + t4) dt ⇒ ∫(t2 + 1) / (1 + t4) dt + ∫(t2 - 1) / (1 + t4) dt ⇒ ∫(1 + 1/t2 ) / (t2 + 1/t2 ) dt + ∫(1 - 1/t2 ) / (t2 + 1/t2 ) dt ⇒ ∫(1 + 1/t2 )dt / [(t - 1/t)2 + 2] + ∫(1 - 1/t2)dt / [(t + 1/t)2 -2] Let t - 1/t = u for the first integral ⇒ (1 + 1/t2 )dt = du and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t2 )dt = dv Integral = ∫du/(u2 + 2) + ∫dv/(v2 - 2) = (1/√2) tan-1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c = (1/√2) tan-1 [(t2 - 1)/t√2] + (1/2√2) log (t2 + 1 - t√2) / t2 + 1 + t√2) + c = (1/√2) tan-1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c
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Shailesh Yadav 8 years, 1 month ago

1
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Rishika Garg 8 years, 1 month ago

1/√1-x^2
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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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Ayush Balyan 8 years, 1 month ago

sin75=sin (30+45)=sin30. cos45+cos30.sin45 =(1+root3)/2.root 2
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Ishu Dagar 8 years, 1 month ago

Anti - derivative. Or togethering of small parts of anything is known as integration
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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}

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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.
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Jaya Maheshwari 8 years, 1 month ago

Then 13 days is enough to prepare..

Jaya Maheshwari 8 years, 1 month ago

If you know all concept In integration then 13 days is enou

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