Solution : Lateral or Curved Surface Area of Cylinder = \(2 \pi rh\) \(\implies\) \(2 \pi rh\) = 94.2 \(\implies\) \(2 \pi r \times 5\) = 94.2 \(\implies\) r = 3 cm Given, height = 5 cm Volume of Cylinder = \(\pi r^2 h\) = \(\pi \times 9 \times 5\) = \(3.14 \times 9 \times …
Solution : Circumference of the base cylindrical vessel = \(2\pi r\) \(\implies\) \(2\pi r\) = 132 \(\implies\) r = 21 cm Given, height = 25 cm Volume of cylinder = \(\pi r^2 h\) = \(\pi \times {21}^2 \times 25\) = 34650 \(cm^3\) Since 1litre = 1000 \(cm^3\) Therefore, It can hold 34.65 litres of …
Solution : We have, I = \(sec^{-1}\sqrt{x}\) dx Let x = \(sec^2t\) dx = \(2sec^2 t tan t\) dt I = t.\(2sec^2 t tan t\) dt u = t and v = \(tan^2 t\) I = \(\int\) u.dv = u.v – \(\int\) v.du = \(t.tan^2 t\) – \(\int\) \(tan^2 t\) dt I = \(t.tan^2 t\) …
Solution : We have, I = \(cos^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(cos^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(cos^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(cos^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(cos^{-1}\sqrt{x}\) – \(\int\) \(-1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(cos^{-1}\sqrt{x}\) – …
Solution : We have, I = \(\int\) \(x cos^{-1} x\) dx By using integration by parts formula, I = \(cos^{-1} x\) \(x^2\over 2\) – \(\int\) \(-1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx I = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx = \(x^2\over 2\) \(cos^{-1} x\) – \(1\over …
Solution : We have, I = \(\int\) \(sin^{-1}(cos x)\) dx By using integration formula, cos x = \(sin({\pi\over 2} – x)\) I = \(\int\) \(sin^{-1}[sin({\pi\over 2} – x)]\) dx I = \(\int\) \(({\pi\over 2} – x)\) dx I = \({\pi\over 2}x\) – \(x^2\over 2\) + C Similar Questions What is the integration of sin inverse …
Solution : We have, I = \(sin^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(sin^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(sin^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(sin^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(sin^{-1}\sqrt{x}\) – \(\int\) \(1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(sin^{-1}\sqrt{x}\) – …
Solution : We have, I = \((sin^{-1}x)^2\) dx Let \(sin^{-1}x\) = t, Then, x = sin t \(\implies\) dx = cos t dt \(\therefore\) I = \(\int\) \((sin^{-1}x)^2\) dx I = \(\int\) \(t^2\) cos t dt Applying integration by parts and, Taking \(t^2\) as first function and cos t as second function, I = \(t^2\) …
What is the integration of sin inverse x whole square ? Read More »
Solution : We have, I = \(\int\) \(x sin^{-1} x\) dx By using integration by parts formula, I = \(sin^{-1} x\) \(x^2\over 2\) – \(\int\) \(1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx I = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over …
Solution : The formula for volume of cuboid of length l, breadth b and height h is \(lbh\). Read Post : Formula for Volume of cuboid – Derivation & Example Similar Questions What is the Formula for Surface Area of Cylinder ? What is the Formula for Surface Area of Cube ? What is the …
