Questions and Answers #4 Analyzing Functions

Questions and Answers Sheet 4 Analyzing Functions Question #1 Derivatives involving other trigonometric functions Find the derivative of the following functions. 𝑦= tan 𝑤 1 + tan 𝑤 Answer: Step 1 d tan(𝑥) ( ) d𝑥 1 + tan(𝑥) 𝑓 ′ Apply the Quotient Rule: ( ) = 𝑓 ′⋅𝑔−𝑔′ ⋅𝑓 𝑔 = 𝑔2 d (tan( x))(1 + tan( x)) dx d (1 + tan( x)) tan( x)(1 + tan( x)) 2 dx = sec 2 (𝑥) (1 + tan(𝑥)) − sec 2 (𝑥) tan(𝑥) (1 + tan(𝑥))2 = 1 1 + sin(2𝑥) Step 2 Question #2 Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f. 𝑓(𝑥) = 2𝑣 𝑖𝑓 𝑥 ≤ 13 − 𝑥 𝑖𝑓 1 < 𝑥 ≤ 4√𝑥 𝑖𝑓 𝑥 > 4 Answer: 2 x , ifx  1 ( x , Ifx  4) we need to find the points of discontinnity. Given the function f ( x) =  3 − x, if1  x  4 Note that 1. f is continnons on (−∞, 1), being an exponential function 2. f is continuous on (1,4), being a polynomial. 3. f is continuous on (4, ∞), being a root function. We only need to determine the continuity at x = 1 and x = 4. Look at the left and right hand limits at those points: lim 𝑓(𝑥) = lim 2𝑥 = 2 𝑥⇒1− 𝑥⇒1− lim 𝑓(𝑥) = lim 3 − 𝑥 = 2 𝑥⇒1+ 𝑥⇒1+ 1 𝑓(𝑥) = 2 = 2 Therefore, f is continuous at x=1 . lim 𝑓(𝑥) lim 3 − 𝑥 = −1 𝑥⇒4− 𝑥⇒4− lim 𝑓(𝑥) lim 𝑥⇒4+ 𝑓(4) = 2 𝑥⇒4+√𝑥 =2 As the left and right hand limits are not same, the functions is discontinuous at x=4 but it is continuous from the right at 4. Question #3 Derivatives involving other trigonometric functions Find the derivative of the following functions. tan 𝑡 𝑦= 1 + sec 𝑡 Answer: To derive the other trigonometric function, 𝑦 = tan 𝑡 1+sec 𝑡 sin 𝑡 cos 𝑡 = 1 1+ cos 𝑡 sin 𝑡 cos 𝑡 = cos 𝑡 + 1 cos 𝑡 Since, 𝑦= sin 𝑡 cos 𝑡 = tan 𝑡 , sec 𝑡 = 1 cos 𝑡 sin 𝑡 1 + cos 𝑡 Differentiate the equation (1) with respect to “x” Apply quotient rules of derivatives, y = d sin t ( ) dt 1 + cos t (1 + cos 𝑡) 𝑑 𝑑 sin 𝑡 − sin 𝑡 (1 + cos 𝑡) 𝑑𝑡 𝑑𝑡 (1 + cos 𝑡)2 = (1 + cos 𝑡) cos 𝑡 − sin 𝑡 (0 − sin 𝑡) (1 + cos 𝑡)2 = cos 𝑡 + cos 2 𝑡 + sin2 𝑡 (1 + cos 𝑡)2 cos 𝑡+1 Further solving it, = (1+cos 𝑡)2 . Since, sin2 𝑡 + cos 2 𝑡 = 1 = 1 + cos 𝑡 + 1 (1 + cos 𝑡)2 𝑦′ = 1 1 + cos 𝑡 Question #4 Derivatives involving other trigonometric functions Find the derivative of the following functions. 𝑦 = csc 2 𝜃 − 1 Answer: Step 1 To differentiate: 𝑦 = csc 2 𝜃 − 1 Solution 𝑦 = csc 2 𝜃 − 1 Differentiating both sides w. r. 𝜃 we get: 𝑑𝑦 𝑑 (csc 2 − 1) = 𝑑𝜃 𝑑𝜃 ⇒ 𝑑𝑦 𝑑 (csc 𝜃) − 0 = (2 csc 𝜃) ⋅ 𝑑𝜃 𝑑𝜃 (𝑢𝑠𝑖𝑛𝑔, ⇒ 𝑑 = (2 csc 𝜃)(− csc 𝜃 cot 𝜃) 𝑑𝜃 (𝑢𝑠𝑖𝑛𝑔, ⇒ 𝑑 (𝑓(𝑔(𝑥)) = 𝑓 ′′ (𝑔(𝑥)) ⋅ 𝑔′ (𝑥))) 𝑑𝑥 𝑑 (csc 𝑥) − cot 𝑥 csc 𝑥) 𝑑𝑥 𝑑𝑦 = −2 csc 2 𝜃 cot 𝑑𝜃 Step 2 Result: 𝑑𝑦 = −2 csc 2 𝜃 cot 𝜃. 𝑑𝜃 Question #5 Derivatives involving other trigonometric functions Find the derivative of the following functions. 𝑦= cot 𝑥 1 + csc 𝑥 Answer: Differentiate the function using quotient rule. 𝑏 𝑑 𝑑𝑦 (1 + csc 𝑥) 𝑑𝑥 (cot 𝑥) − (cot 𝑥) 𝑑𝑥 (1 + csc 𝑥) = (1 + csc 𝑥)2 𝑑𝑥 Simplify further: 𝑦′ = = (− csc 𝑥)2 (1 + csc 𝑥) − (− cot 𝑥 csc 𝑥)(cot 𝑥) (1 + csc 𝑥)2 − csc 2(𝑥) (1 + csc 𝑥) + cot 2 𝑥 csc 𝑥 (1 + csc 𝑥)2 Therefore, 𝑦 ′ = − csc2 (𝑥)(1+csc 𝑥)+cot2 𝑥 csc 𝑥 (1+csc 𝑥)2 Question #6 Average and marginal cost Consider the following cost functions. a. Find the average cost and marginal cost functions. b. Determine the average cost and the marginal cost when 𝑥 = 𝑎. c. Interpret the values obtained in part (b). 𝐶(𝑥) = 500 + 0.02𝑥, 0 ≤ 𝑥 ≤ 2000, 𝑎 = 1000 Answer: Step1 Given: Cost function 𝐶(𝑥) = 500 + 0.02𝑥, 0 ≤ 𝑥 ≤ 2000 a. The average cost = 𝐶𝑎𝑣 (𝑥) = average cos 𝑡 𝐶𝑎𝑣 (𝑥) = 500 𝑥 𝐶(𝑥) 𝑥 = 500+0.02𝑥 𝑥 + 0.02 The marginal cos 𝑡 = 𝑀(𝑥) = 𝑑 𝑑𝑥 𝐶(𝑥) = 𝐶 ′ (𝑥) = 0.02 Step 2 b. The average cos 𝑡 = 𝐶𝑎𝑣 (𝑥) = 500+0.02𝑥 𝑥 average cos 𝑡 for 𝑥 = 1000 = 𝐶𝑎𝑣 (1000) = 500+0.02(1000) 1000 = 0.52 The marginal cos 𝑡 = 𝑀(𝑥) = 0.02𝑤 The marginal 𝑐𝑜𝑠 t for 𝑥 = 1000 is 𝑀(1000) = 0.02 c. The marginal cost for x = 1000 is 0.02, whereas the average cost is 0.52. The average cost is higher because of the fixed cost of 500 in the cost function. The marginal cost does not take into account these fixed cost. It is only a measure of the incremental cost. Question #7 Determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell why not. 𝐹(𝑥) = 5𝑥 4 − 𝜋𝑥 3 + 1 2 Answer: Step 1 Given function is 𝐹(𝑥) = 5𝑥 4 − 𝜋𝑥 3 + 1 2 To determine whether functions are polynomial functions. Solution: A polynomial function is a function which have only non-negative integers power of variables. Given function is 𝐹(𝑥) = 5𝑥 4 − 𝜋𝑥 3 + 1 2 Here, we can find that power of variable in non-negative integers. Therefore, given function isa polynomial function. Step 2 The degree of a polynomial is the highest power of variables in the expression. In the given function, the highest power of the variable is 4. Therefore, degree of the polynomial is 4. Hence, given function isa polynomial function and its degree is 4. Question #8 Determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell why not. 𝐺(𝑥) = −3𝑥 2 (𝑥 + 2)3 Answer: Step 1 We know, A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer powers of variables. Given, 𝑓(𝑥) = −3𝑥 2 (𝑥 − 2)3 𝑓(𝑥) = −3𝑥 2 (𝑥 3 + 3𝑥 2 (2) + 3𝑥(2)2 + 23 ) 𝑓(𝑥) = −3𝑥 2 (𝑥 3 + 6𝑥 2 + 12𝑥 + 8) 𝑓(𝑥) = −3𝑥 5 − 18𝑥 4 − 36𝑥 3 − 24𝑥 2 Here, this is Sth degree of the polynomial. Step 2 Therefore, 𝑓(𝑥) = −3𝑥 5 − 18𝑥 4 − 36𝑥 3 − 24𝑥 2 is 5th degree of the polynomial. Question #9 𝑦 = 𝑒 𝑥 sin ℎ 𝑥 Answer: Step 1 𝑦 = 𝑒 (𝑥)sin ℎ (𝑥) Apply product rule for taking derivative 𝑦 = 𝑢𝑣 𝑦 ′ = 𝑢′ 𝑣 + 𝑣 ′ (𝑢) derivative of sin ℎ (𝑥) is cos ℎ (𝑥) Step 2 𝑦 = 𝑒 𝑥 sin ℎ (𝑥) 𝑦 ′ = 𝑒 𝑥 sin ℎ (𝑥) + 𝑒 𝑥 cos ℎ (𝑥) 𝑦 ′ = 𝑒 𝑥 (𝑠𝑖𝑛 ℎ(𝑥) + cos ℎ (𝑥)) Question #10 ∫ 2𝑒 −5𝑥 𝑑𝑥 ∫ 𝑒 8𝑥 𝑑𝑥 Answer: Step 1 𝐼 = ∫ 2𝑒 −5𝑥 𝑑𝑥 −5𝑥 = 𝑡 −5𝑑𝑥 = 𝑑𝑥 −1 𝐼 = 2 ( ) ∫ 𝑒 𝑡 𝑑𝑡 5 𝐼= −2 ∫ 𝑒 𝑡 𝑑𝑡 5 𝐼= −2 𝑡 𝑒 +𝑐 5 𝐼= −2 −5𝑥 𝑒 +𝑐 5 Step 2 𝐼 = ∫ 𝑒 8𝑥 𝑑𝑥 8𝑥 = 𝑡 8𝑑𝑥 = 𝑑𝑡 𝐼= 1 ∫ 𝑒 𝑡 𝑑𝑡 8 𝐼= 1 𝑡 𝑒 +𝑐 8 𝐼= 1 8𝑥 𝑒 +𝑐 8 Question #11 Critical Numbers Use a graphing utility to graph the following four functions. Only one of the functions has critical numbers. Which is it? Explain. 𝑓(𝑥) = 𝑒 𝑥 Answer: Step 1 Given a function 𝑓(𝑥) = 𝑒 𝑥 To find whether the function has critical numbers or not Equate the first derivative 𝑓 ′ (𝑥) = 0 ⇒ 𝑒𝑥 = 0 There are no values of x that make this equation true Hence, it has no solution Step 2 Hence, the given function does not have critical number Question #12 Derivatives of power and constant functions Find the derivative of the following functions. 𝑦 = 𝑥5 Answer: Find 𝑑 𝑑𝑥 (𝑥 5 ) Apply the power rule ( 𝑑 𝑑𝑥 (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 with 𝑛 = 5: d 5 ( x )) = (5 x 4 ) dx Thus, 𝑑 𝑑𝑥 (𝑥 5 ) = 5𝑥 4 Answer: 𝑑 5 (𝑥 ) = 5𝑥 4 𝑑𝑥 Question #13 Determine which functions are polynomial functions. For those that are, identify the degree. 𝑓(𝑥) = (𝑥 2 + 7)/3 Answer: Step 1 We know, A polynomial function is a function such as a quadratic, a cubic, a quartic, and so on, involving only non-negative integer powers of and degree of a polynomial is the highest power of the involved variable, Now, given, 𝑓(𝑥) = 𝑥 2 +7 3 1 7 3 3 ⇒ 𝑓(𝑥) = 𝑥 2 + The given function is a polynomial as it contains non-negative integral power of x. And highest power of variable involved (x) is 2 Hence, degree=2 Step 2 Result: Given function is a polynomial function. And degree=2 Question #14 Determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell why not. 1 ℎ(𝑥) = 3 − 𝑥 2 Answer: Step1 The polynomial expression consists of variables and their coefficients. The variables are presented in the expression with exponents. The general form of polynomial equation of degree n is 𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑝𝑖” ′+ + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0 . The degree of a polynomial is the highest power of variable in the polynomial equation. A function is called a polynomial function if it can be expressed in the form of a polynomial expression. The solution of a polynomial is the value for the variable (x) that satisfies the polynomial equation. Step 2 1 Consider the equation ℎ(𝑥) = 3 − 𝑥 . 2 1 A polynomial function involves only positive integer powers forts variable. Here, in ℎ(𝑥) = 3 − 𝑥, the 2 variable is x and its power is non negative ( power equal to 1). 1 Hence, the given equation ℎ(𝑥) = 3 − 𝑥 is a polynomial function, which is of the form of a polynomial 2 with degree one as 𝑃(𝑥) = 𝑎1 𝑥 = 𝑎[0} . The degree of a polynomial equation is the highest variable power of the polynomial. 1 Here, we have the highest power of polynomial ℎ(𝑥) = 3 − 𝑥 is 1 . 2 Hence, the function is a polynomial function with degree 1. Question #15 Sketch the graphs of the functions. f(x) =x^{3}+x^{2}- 2x\ and\ g(x) =0 Find the ares of the rezion completely enclosed by the graphs of the given functions fand g. Answer: Step 1 Sketch the region: Step 2 𝑏 Ares between the curves, 𝐴 = ∫𝑎 |𝑓(𝑥) − 𝑔(𝑥)|𝑑𝑥 Equating two functions, 𝑋 3 + 𝑥 2 − 2𝑥 = 0 Solve, 𝑥 = −2, 𝑥 = 1 . 1 𝐴 = ∫ |𝑥 3 + 𝑥 2 − 2|𝑑𝑥 −2 0 1 Eliminate absolute, 𝐴 = ∫−2 𝑥 3 + 𝑥 2 − 2𝑥𝑑𝑥 + ∫0 − 𝑥 3 − 𝑥 2 + 2𝑥𝑑𝑥 8 5 3 12 Now integrate the term, 𝐴 = + = 37 12 Thus area between the curve is 𝐴 = 37 12 Question #16 Use function composition to determine if f (x) and g (x) are inverse functions. 3 𝑓(𝑥) = √𝑥 + 1 and 𝑔(𝑥) = 𝑥 3 + 1 Answer: 3 Viven functions 𝑓(𝑥) = √𝑥 + 1, 𝑔(𝑥) = 𝑥 3 + 1 we know that, using composition functions If f(x) and g(x) are invevsc functions then. 𝑓(𝑔(𝑥)) = 𝑥, 𝑔(𝑓(𝑥)) = 𝑥 Checking 𝑓(𝑔(𝑥)) = 𝑓(𝑥 3 + 1) 3 = √(𝑥 3 + 1) + 1 3 𝑓(𝑔(𝑥)) = √𝑥 + 2 ≠ 𝑥. 3 𝑔(𝑓(𝑥)) = 𝑔( √𝑥 + 1) 3 3 = ( √𝑥 + 1) + 1 = (𝑥 + 1) + 1 𝑔(𝑓(𝑥)) = 𝑥 + 2 ≠ 𝑥 𝑔(𝑓(𝑥)) = 𝑓(𝑔(𝑥)) ≠ 𝑥 So,they ave not invevsc functions Question #17 A(t) is a 2 × 2 matrix of differentiable functions and X(t) is a 2 × 1 column matrix of differentiable functions, prove the product rule Answer Step 1 Let The matrices A(t) and X(t) be  A (t ) A(t ) =  11  A21 (t ) A12 (t )   x1 (t ) X ( t ) =  x (t ) A22 (t )   2    The product rule of differentiation states that 𝑑 [𝐴(𝑡)𝑋(𝑡)] = 𝐴′ (𝑡)𝑋(𝑡) + 𝐴(𝑡)𝑋 ′ (𝑡) 𝑑𝑡 Step 2  A (t ) A(t ) X (t ) =  11  A21 (t )  A (t ) x1 (t ) =  11  A21 (t ) x1 (t ) Now, 𝑑 𝑑𝑡 A12 (t )   x1 (t ) A22 (t )   x2 (t )    A12 (t ) x2 (t )  A22 (t ) x2 (t )  [𝐴(𝑡)𝑋(𝑡)] d  dt A11 (t ) x1 (t ) =  d A (t ) x (t )  dt 21 1  A12 (t ) x2 (t )   A22 (t ) x2 (t )   d  dt [ A11 (t ) x1 (t ) =  d [ A (t ) x (t ) 1  dt 21  A12 (t ) x2 (t )]  A22 (t ) x2 (t )]   A  (t ) x1 (t ) + A11 (t ) x1 (t ) + =  11  A21 (t ) x1 (t ) + A21 (t ) x1 (t ) + A12 (t ) x2 (t ) + A12 (t ) x2 (t )  A22 (t ) x2 (t ) + A22 (t ) x2 (t )  Step 3 Now,  A  (t ) A(t ) =  11  A21 (t )  x  (t ) X (t ) =  1  x2 (t ) Therefore, A12 (t )  A22 (t )      A  (t ) A(t ) X (t ) + A(t ) X (t ) =  11  A21 (t )  A  (t ) x1 (t ) + =  11  A21 (t ) x1 (t ) + A12 (t ) x2 (t )   A11 (t ) x1 (t ) + + A22 (t ) x2 (t )   A21 (t ) x1 (t ) +  A  (t ) x1 (t ) + A11 (t ) x1 (t ) + =  11  A21 (t ) x1 (t ) + A21 (t ) x1 (t ) + Thus, A12 (t )   x1 (t ) A22 (t )   x2 (t )   A11 (t )  +  A (t )   21 A12 (t )   x1 (t ) A22 (t )   x2 (t )    A12 (t ) x2 (t )  A22 (t ) x2 (t )  A12 (t ) x2 (t ) + A12 (t ) x2 (t )  A22 (t ) x2 (t ) + A22 (t ) x2 (t )  d [ A(t ) X (t )] = A(t ) X (t ) + A(t ) X (t ) dt Question #18 Find the first derivative of the following functions and inverse trigonometric functions. Use the formulas 19. 𝜃 = arccos √1 − 𝑥 Answer: Given 𝜃 = arccos √1 − 𝑥 ′ −1 Apply differentiation nale [arccos(𝑢(𝑥)) = √1−(𝑢(𝑥)) 1 2 𝑢′ (𝑥) have 𝑢(𝑥) = √1 − 𝑥 𝑢′ (𝑥) = 1 Agply, power ale (𝑓(𝑥)𝑛 )′ = 𝑛𝑓(𝑥)𝑛−1 . f'(x) 𝑢′ (𝑥 (1 − 𝑥)2−1 ) 2 𝑑 𝑑𝑥 𝑑 𝑑𝑥 1 (1 − 𝑥)2 (1 − 𝑥) Apply (𝑓 ± 9)′ = 𝑓′ ± 9 (𝑎𝑛 )′ = 𝑛𝑥 𝑛−1 , 𝑢′ (𝑥) = 𝑑 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) = 0 𝑑𝑥 1 2√1 − 𝑥 (0 − 1) = [(arccos √1 − 𝑥 ′ = −1 2√1 − 𝑥 1  −1  d   dx = 2 1− x x  2 1− x  −1 √1−(1−𝑥) Question #19 Use this only Power Formula Logarithm Formula Exponential Functions Trigonometric Functions. Complete solution 1 ∫ 0 2𝑥 3 + 𝑥 2 + 3 𝑑𝑥 (𝑥 2 + 1)2 Answer: 1 2𝑥 3 +𝑥 2 +3 Given ∫0 (𝑥 2 +1)2 𝑑𝑥 Perform partial decomposition: =∫ 2𝑥 + 1 2 − 2𝑥 + 2 𝑑𝑥 2 𝑥 + 1 (𝑥 + 1)2 Apply lienarity: =∫ 2𝑥 + 1 𝑥−1 𝑑𝑥 − 2 ∫ 2 𝑑𝑥 (𝑥 + 1)2 𝑥2 + 1 2𝑥+1 Now solve ∫ 2 𝑑𝑥 𝑥 +1 Expand: =∫ 2𝑥 1 𝑑𝑥 + ∫ 2 𝑑𝑥 +1 𝑥 +1 𝑥2 Substitute 𝑢 = 𝑥 2 + 1 ⇒ 𝑑𝑢 𝑑𝑥 = 2𝑥 ⇒ 𝑑𝑥 = 1 1 = ∫ 𝑑𝑢 2 𝑢 1 We knov that ∫ 𝑑𝑥 = 𝐼𝑛(𝑥) 𝑥 = ln(𝑢) 2 Undo the substitution: = 𝐼𝑛(𝑥 2 + 1) 2 1 2𝑥 𝑑𝑢 1 We know that ∫ 2 𝑑𝑥 = arctan(𝑥) 𝑥 +1 Plug in solved integrals: = 𝐼𝑛(𝑥 2 + 1)+∣ arctan(𝑥) 𝑥−1 Now solvind: ∫ (𝑥 2 2 +1) Expand: =∫ (𝑥 2 𝑥 1 𝑑𝑥 − ∫ 2 𝑑𝑥 2 (𝑥 + 1)2 + 1) Substitute 𝑢 = 𝑥 2 + 1 ⇒ 𝑑𝑢 𝑑𝑥 = 2𝑥 ⇒ 𝑑𝑥 = 1 2𝑥 𝑑𝑢 1 1 = ∫ 2 𝑑𝑢 2 𝑢 We know that ∫ 𝑢𝑛 𝑑𝑢 = =− 𝑢𝑛+1 𝑛+1 1 2𝑢 Undo the substitute = − 1 2(𝑥 2 +1) Apply reduction formula: ∫ 1 2𝑛 − 3 1 𝑥 𝑑𝑥 = ∫ 𝑑𝑥 + (𝑎𝑥 2 + 𝑏)𝑛 2𝑏(𝑛 − 1) (𝑎𝑥 2 + 𝑏)𝑛−1 2𝑏(𝑛 − 1)(𝑎𝑥 2 + 𝑏)𝑛−1 With a=1,b=2 and n=2: = = 𝑥 1 1 + ∫ 2 𝑑𝑥 + 1) 2 𝑥 + 1 2(𝑥 2 arctan(𝑥) 𝑥 + 2 2(𝑥 2 + 1) Plug in the solved integrals: = 𝐼𝑛(𝑥 2 + 1) + 2 arctan(𝑥) + 𝑥+1 +𝐶 𝑥2 + 1 Now substitute the limits: [=( In(1 + 1) + 2 arctan(1) + = 0 +1 1+1 )) ] )-( In(0 + 10 + 2 arctan(0) + 0 +1 1+1 2𝐼𝑛(2) + 𝜋 + 2 −1 2 = 2.2639 1 2𝑥 3 +𝑥 2 +3 Therefore ∫0 (𝑥 2 +1)2 Question #20 1. ∫ sin 2 3 𝑥𝑑𝑥 𝑥 2. ∫ cos 6 𝑥 2 𝑑𝑥 3. ∫ tan 9 𝑥𝑑𝑥 𝑑𝑥 = 2.2639 4. ∫ 𝑥 2 cot 𝑥 3 𝑑𝑥 Answer: Solution of question 1 𝑠 2 ∫ 𝑖𝑛 ( 𝑥) 𝑑𝑥 3 Put 𝑢 = 2𝑥 3 ⇒ 𝑑𝑢 𝑑𝑥 2 3𝑑𝑢 3 2 = or = 𝑑𝑥 3 3 ⇒ ∫ sin 𝑢 𝑑𝑢 = ∫ sin 𝑢 𝑑𝑢 2 2 Now solving integral = 3 − cos(𝑢) + 𝑐 2 = −3 cos 𝑢 +𝑐 2 = −3 cos ( 2 2 ) 3𝑥 + 𝑐 2 Put back 𝑢 = ( ) 𝑥 3 Hence above is the required solution pf given integral Solution of question 3 ∫ × cos(6𝑥 2 ) 𝑑𝑥 Put 𝑢 = 6𝑥 2 ⇒ ⇒ ∫ 𝑥 cos 𝑢 ⇒ 𝑑𝑢 𝑑𝑥 = 12𝑥 ⇒ 𝑑𝑥 = 1 12𝑥 𝑑𝑢 1 𝑑𝑢 12𝑥 1 ∫ cos 𝑢 𝑑𝑢 12 Now solving integral ⇒ 1 12 sin(𝑢) + 𝑐 Put back 𝑢 = 6𝑥 2 ⇒ sin 6 𝑥 2 +𝑐 12 Hence above is the required solution of given integral Solution of question 3 ∫ tan 9 𝑥 Put 𝑢 = 9𝑥 ⇒ ⇒ 𝑑𝑢 𝑑𝑥 1 = 0 ⇒ 𝑑𝑢 = 𝑑𝑢 9 1 ∫ tan(𝑢) 𝑑𝑥 0 Now solving integral Rewosite tan(𝑢) = sin 𝑢 cos 𝑢 ⇒∫ 1 sin 𝑢 𝑑𝑢 9 cos 𝑢 Substitute 𝑣 = cos 𝑢 𝑑𝑣 = − sin 𝑢 𝑑𝑢 = ⇒ 𝑑𝑢 = − ⇒ 1 𝑑𝑣 sin 𝑢 1 sin 𝑢 (−1) ∫ ⋅ 𝑑𝑣 9 𝑣 sin 𝑢 1 1 ⇒ − ∫ 𝑑𝑣 9 𝑣 Now solving integral = − log(𝑣) /9 + 𝑐 Put back 𝑣 = cos 𝑢 ⇒ − log(cos 𝑢) +𝑐 9 Put back u=9x ⇒ − log(cos(9𝑥)) +𝑐 9 ⇒ log(cos(9𝑥))−1 +𝑐 9 ⇒ 1 log(sec(9𝑥)) + 𝑐 9 Hence above is the required solution of given integral Solution of question 4 𝑥 2 cot (𝑥 2 ) 𝑑𝑥 ∫ Put 𝑢 = 𝑥 3 ⇒ 𝑥 2 cot =∫ 𝑢⋅ 𝑑𝑢 𝑑𝑥 = 3𝑥 2 ⇒ 3𝑥 2 ⇒ 𝑑𝑥 = 1 𝑑𝑥 3𝑥 2 1 ∫ cot 𝑢 𝑑𝑥 3 Now Solving integral 1 cos(𝑢) = ∫ 𝑑𝑢 3 sin(𝑢) Put 𝑣 = sin(𝑢) ⇒ ⇒ 𝑑𝑢 = ⇒∫ 𝑑𝑣 𝑑𝑢 = cos 𝑢 1 𝑑𝑣 cos 𝑢 1 cos(𝑢) 1 ⋅ 𝑑𝑢 3 𝑣 cos 𝑢 1 3𝑥 2 𝑑𝑢 1 1 ⇒ ∫ ∫ 𝑑𝑣 3 𝑣 Now solving integral ⇒ 1 log(𝑣) + 𝑐 3 Put back 𝑣 = sin 𝑢 ⇒ 1 log(sin 𝑢) + 𝑐 3 again put back 𝑢 = 𝑥 3 ⇒ 1 log(sin 𝑥 2 ) + 𝑐 3 Hence aboveis the required of given integral