Nettet31. mai 2024 · 1+tan 2 x = sec 2 x. I = ∫sec 2 x sec 4 x dx = ∫sec 2 x (1 + tan 2 x) 2 dx . Now, ... Now put the value of t, which is t = tan x in above integral-I = tanx + \(\frac{tan^5x}{5}\) + 2. \(\frac{tan^3x}{3}\) ← Prev Question Next ... Evaluate ∫ tan^3x sec^4x dx. asked May 31, 2024 in Indefinite Integral by rahul01 (29.3k points) Nettet7. sep. 2024 · In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals.They are an …
integral of sec^5x-sec^3x - Symbolab
Nettet14. jan. 2024 · Explanation: Split up tan3(x) into tan2(x)tan(x) then rewrite tan2(x) using the identity tan2(θ) +1 = sec2(θ) ⇒ tan2(θ) = sec2(θ) − 1. ∫tan3(x)dx = ∫tan2(x)tan(x)dx = ∫(sec2(x) −1)tan(x)dx Distribute: = ∫sec2(x)tan(x)dx −∫tan(x)dx For the first integral, apply the substitution u = tan(x) ⇒ du = sec2(x)dx, both of which are already in the integral. NettetSolution Verified by Toppr Correct option is C) ∫sec 2(3x+5)dx let 3x+5=t 3dx=dt dx= 1/ 3dt ∫sec 2t⋅ 31⋅dt = dfrac13∫sec 2t⋅dt = 31(tant)+c = 31(tan(3x+5))+c ∫sec 2(3x+5)dx= 31(tan(3x+5))+c Solve any question of Integrals with:- > Was this answer helpful? 0 0 ∫x 6+13x 2 dx Medium View solution ∫e xsece xtane xdx= Medium Integrals 9 10 8 10 firefox the movie
Evaluate the Integral integral of sec(3x)tan(3x) with respect to x ...
NettetCalculus Evaluate the Integral integral of tan (x)^5sec (x)^7 with respect to x ∫tan 5(x)sec 7(x)dx Factor out tan4(x). ∫tan4(x)tan(x)sec7(x)dx Simplify with factoring out. Tap for … Nettet17. nov. 2024 · answered Nov 17, 2024 by Jay01 (39.7k points) selected Nov 18, 2024 by Vikash Kumar. Best answer. Let I = ∫ sec3 x dx. = ∫ (sec2 x .sec x) dx. = sec x ∫ sec2 x dx – ∫ (sec x .tan x .tan x) dx. = sec x ∫ sec2 x dx – ∫ (sec x .tan2 x) dx. = sec x. tan x – ∫ (sec x (sec2 x – 1)) dx. = sec x. tan x – ∫ sec3 x dx + ∫ sec ... Nettet14. sep. 2012 · Calculate the following integral: ∫ sec^4 (3x)/ tan^3 (3x) dx For this one, can I bring up the tan to tan^-3? asked by Robert September 14, 2012 2 answers yes, so you have ∫sec^4 (3x)* tan^-3 (3x) dx ∫sec^2 (3x)* (tan^2 (3x)+1)*tan^-3 (3x) dx Now let u = tan (3x) du = 3sec^2 (3x) dx ∫1/u + 1/u^3 du/3 = 1/3 ln (u) - 1/6 u^-2 ethenoanthracene