Q. 75 Prove聽(a)聽鈭�11-x2聽dx=sin-1聽... [FREE SOLUTION]

Chapter 4: Q. 75 (page 363)

Prove
$(a) 鈭� \frac{1}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x d x + C using differentiation formulas (b) 鈭� \frac{1}{1 + x^{2}} d x = \tan^{- 1} x d x + C' (c) 鈭� \frac{1}{|x| \sqrt{x^{2} - 1}} d x = s e c^{- 1} x d x + C$

Short Answer

Expert verified

Hence Proved

Step by step solution

Step 1. To prove

$(a) 鈭� \frac{1}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x d x + C using differentiation formulas (b) 鈭� \frac{1}{1 + x^{2}} d x = \tan^{- 1} x d x + C' (c) 鈭� \frac{1}{|x| \sqrt{x^{2} - 1}} d x = s e c^{- 1} x d x + C$

Part(a) Step 2. Proving

$(a) 鈭� \frac{1}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x d x + C using differentiation formulas \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}} so, \sin^{- 1} x is the antiderivative of \frac{1}{\sqrt{1 - x^{2}}} Hence Proved$

Part(b) Step 3. Proving

$(b) 鈭� \frac{1}{1 + x^{2}} d x = \tan^{- 1} x d x + C using differentiation formulas \frac{d}{d x} (\tan^{- 1} x) = \frac{1}{1 + x^{2}} so, \tan^{- 1} x is the antiderivative of \frac{1}{1 + x^{2}} Hence Proved$

Part (c) Step 4. Proving

$(c) 鈭� \frac{1}{|x| \sqrt{x^{2} - 1}} d x = s e c^{- 1} x d x + C using differentiation formulas \frac{d}{d x} (s e c^{- 1} x) = \frac{1}{|x| \sqrt{x^{2} - 1}} so, \sec^{- 1} x is the antiderivative of \frac{1}{|x| \sqrt{x^{2} - 1}} Hence Proved$

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91影视

Prove
$(a) 鈭� \frac{1}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x d x + C using differentiation formulas (b) 鈭� \frac{1}{1 + x^{2}} d x = \tan^{- 1} x d x + C' (c) 鈭� \frac{1}{|x| \sqrt{x^{2} - 1}} d x = s e c^{- 1} x d x + C$

Short Answer

Step by step solution

Step 1. To prove

Part(a) Step 2. Proving

Part(b) Step 3. Proving

Part (c) Step 4. Proving

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Prove(a)鈭�11-x2dx=sin-1xdx+Cusingdifferentiationformulas(b)鈭�11+x2dx=tan-1xdx+C'(c)鈭�1xx2-1dx=sec-1xdx+C

Short Answer

Step by step solution

Step 1. To prove

Part(a) Step 2. Proving

Part(b) Step 3. Proving

Part (c) Step 4. Proving

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Statistics

Discrete Mathematics

Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

Prove
$(a) 鈭� \frac{1}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x d x + C using differentiation formulas (b) 鈭� \frac{1}{1 + x^{2}} d x = \tan^{- 1} x d x + C' (c) 鈭� \frac{1}{|x| \sqrt{x^{2} - 1}} d x = s e c^{- 1} x d x + C$